Translational Motion - MCAT Physical
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A 2kg lead ball is loaded into a spring cannon and the cannon is set at a 45º angle on a platform. The spring has a spring constant of 100N/m and the ball and spring system is compressed by 1m before launch. While the ball is in flight air resistance can be neglected, and the ball finishes its flight by landing at a cushion placed some distance away from the cannon.
Find the horizontal component of velocity once the ball has left the cannon.
A 2kg lead ball is loaded into a spring cannon and the cannon is set at a 45º angle on a platform. The spring has a spring constant of 100N/m and the ball and spring system is compressed by 1m before launch. While the ball is in flight air resistance can be neglected, and the ball finishes its flight by landing at a cushion placed some distance away from the cannon.
Find the horizontal component of velocity once the ball has left the cannon.
This asks us to understand the vector components of velocity. Remember that the final velocity is the hypotenuse of a triangle (solved to be 10m/s in the previous problem), and that by knowing the hypotenuse value we can solve for the horizontal component by using cosine.
vx = (10m/s)(cos(45o)) = 7.1m/s
This asks us to understand the vector components of velocity. Remember that the final velocity is the hypotenuse of a triangle (solved to be 10m/s in the previous problem), and that by knowing the hypotenuse value we can solve for the horizontal component by using cosine.
vx = (10m/s)(cos(45o)) = 7.1m/s
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A 2kg lead ball is loaded into a spring cannon and the cannon is set at a 45º angle on a platform. The spring has a spring constant of 100N/m and the ball and spring system is compressed by 1m before launch. While the ball is in flight air resistance can be neglected, and the ball finishes its flight by landing at a cushion placed some distance away from the cannon.
What is the initial vertical component of velocity of the ball?
A 2kg lead ball is loaded into a spring cannon and the cannon is set at a 45º angle on a platform. The spring has a spring constant of 100N/m and the ball and spring system is compressed by 1m before launch. While the ball is in flight air resistance can be neglected, and the ball finishes its flight by landing at a cushion placed some distance away from the cannon.
What is the initial vertical component of velocity of the ball?
This question also asks us to understand the vector components of velocity. Remember that the final velocity is the hypotenuse of a triangle, and that by knowing the hypotenuse value (solved as 10m/s in a previous problem) we can solve for the vertical component using sine.
vy = (10m/s)(sin(45o)) = 7.1m/s
This question also asks us to understand the vector components of velocity. Remember that the final velocity is the hypotenuse of a triangle, and that by knowing the hypotenuse value (solved as 10m/s in a previous problem) we can solve for the vertical component using sine.
vy = (10m/s)(sin(45o)) = 7.1m/s
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A 2kg lead ball is loaded into a spring cannon and the cannon is set at a 45º angle on a platform. The spring has a spring constant of 100N/m and the ball and spring system is compressed by 1m before launch. While the ball is in flight air resistance can be neglected, and the ball finishes its flight by landing at a cushion placed some distance away from the cannon.
What is the horizontal acceleration of the ball during its flight?
A 2kg lead ball is loaded into a spring cannon and the cannon is set at a 45º angle on a platform. The spring has a spring constant of 100N/m and the ball and spring system is compressed by 1m before launch. While the ball is in flight air resistance can be neglected, and the ball finishes its flight by landing at a cushion placed some distance away from the cannon.
What is the horizontal acceleration of the ball during its flight?
Without an additional force acting in the horizontal direction during flight (we are told we can neglect air resistance), there is no acceleration. Remember that Newton's second law, F = ma, requires that a force act on an object to produce acceleration. Here, we have no additional force and thus no acceleration.
Without an additional force acting in the horizontal direction during flight (we are told we can neglect air resistance), there is no acceleration. Remember that Newton's second law, F = ma, requires that a force act on an object to produce acceleration. Here, we have no additional force and thus no acceleration.
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In which of the following collisions is momentum not conserved?
In which of the following collisions is momentum not conserved?
Momentum is conserved in both elastic and inelastic collisions. During an elastic collision, kinetic energy is also conserved and the two objects remain separate after impact. During an inelastic collision, kinetic energy is lost to the surroundings and the objects traditionally stick together upon impact. The energy lost in an inelastic collision is generally transformed into heat or sound.
The only time momentum will not be conserved is when an outside force is introduced or the masses of objects do not remain constant.
Momentum is conserved in both elastic and inelastic collisions. During an elastic collision, kinetic energy is also conserved and the two objects remain separate after impact. During an inelastic collision, kinetic energy is lost to the surroundings and the objects traditionally stick together upon impact. The energy lost in an inelastic collision is generally transformed into heat or sound.
The only time momentum will not be conserved is when an outside force is introduced or the masses of objects do not remain constant.
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Suppose a ball with mass of 10kg was dropped from rest from the top of a cliff that is 80m tall. How long will it take the ball to reach the bottom of the cliff?
Suppose a ball with mass of 10kg was dropped from rest from the top of a cliff that is 80m tall. How long will it take the ball to reach the bottom of the cliff?
The question tells us that the initial velocity of the ball (v0) is zero and the height (d) of the cliff is 80m. Acceleration (gravity) is 10m/s2. Using the kinematics eqation 
we can solve for time.
80 = 0 + (1/2)(10)t2
80/5 = t2 = 16
t = 4s
The question tells us that the initial velocity of the ball (v0) is zero and the height (d) of the cliff is 80m. Acceleration (gravity) is 10m/s2. Using the kinematics eqation
80 = 0 + (1/2)(10)t2
80/5 = t2 = 16
t = 4s
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If a 15kg ball takes five seconds to strike the ground when released from rest, at what height was the ball dropped?
If a 15kg ball takes five seconds to strike the ground when released from rest, at what height was the ball dropped?
Using the equation 
we can find the distance at which the ball was dropped. Notice that the mass of the ball does not matter in this problem. We are told that the ball is dropped from rest making, 
, thus we have 
. When we plug in our values, and assuming that acceleration is equal to gravity (10m/s2) we find that 
= 125m.
Using the equation
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Ball A, traveling 
to the right, collides with ball B, traveling 
to the left. If ball A is 4kg and ball B is 6kg, what will be the final velocity and direction after a perfectly inelastic collision?
Ball A, traveling
A perfectly inelastic collision is when the two bodies stick together at the end. At the beginning the two balls are traveling separately with individual momentum values. Using the momentum equation 
, we can see that ball A has a momentum of (4kg)(7m/s) to the right and ball B has a momentum of (6kg)(8m/s) to the left. The final momentum would be the mass of both balls times the final velocity, (4+6)(vf). We can solve for vf through conservation of momentum; the sum of the initial momentum values must equal the final momentum.
Note: ball B's velocity is negative because they are traveling in opposite directions.
The negative sign indicates the direction in which the two balls are traveling. Since the sign is negative and we indicated that traveling to the left is negative, the two balls must be traveling 2m/s to the left after the perfectly inelastic collision.
A perfectly inelastic collision is when the two bodies stick together at the end. At the beginning the two balls are traveling separately with individual momentum values. Using the momentum equation
Note: ball B's velocity is negative because they are traveling in opposite directions.
The negative sign indicates the direction in which the two balls are traveling. Since the sign is negative and we indicated that traveling to the left is negative, the two balls must be traveling 2m/s to the left after the perfectly inelastic collision.
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An object of mass 
is moving at a velocity of 
when it undergoes a perfect inelastic collision with a stationary mass. After the collision, the objects are moving at velocity of 
. What is the mass of the stationary object?
An object of mass
Momentum is given by the equation 
. Conservation of momentum states that 
.
The initial momentum is given by the sum of the initial momentums of the two objects. Because the second mass starts from rest, it becomes irrelevant in the equation.
After a perfectly inelastic collision, the two objects will stick together and their masses will add. The final velocity will apply to the combined mass of the two objects.
Since we know that 
, we can use our values from these calculations to solve for 
.
The mass of the second object must be 2kg.
Momentum is given by the equation
The initial momentum is given by the sum of the initial momentums of the two objects. Because the second mass starts from rest, it becomes irrelevant in the equation.
After a perfectly inelastic collision, the two objects will stick together and their masses will add. The final velocity will apply to the combined mass of the two objects.
Since we know that
The mass of the second object must be 2kg.
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A 200-gram object moves in the +x direction at 4m/s and collides with an identical object moving in the –y direction at 3m/s. If the two objects stick together after the collision, what is the magnitude of their resulting velocity?
A 200-gram object moves in the +x direction at 4m/s and collides with an identical object moving in the –y direction at 3m/s. If the two objects stick together after the collision, what is the magnitude of their resulting velocity?
Use conservation of momentum in each direction: 
and 
The initial momentum in the x direction is provided only by the first object.
This also equals the final x-momentum, when the two objects move together at the same velocity.
The initial momentum in the y direction is provided only by the second object.
Again, this equals the final y-momentum.
Combine the two velocities using the Pythagorean theorem.
Use conservation of momentum in each direction:
The initial momentum in the x direction is provided only by the first object.
This also equals the final x-momentum, when the two objects move together at the same velocity.
The initial momentum in the y direction is provided only by the second object.
Again, this equals the final y-momentum.
Combine the two velocities using the Pythagorean theorem.
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Which of the following is a correct description of an inelastic collision?
Which of the following is a correct description of an inelastic collision?
Momentum is conserved in any collision, but kinetic energy is only conserved in elastic collisions. So, an inelastic collision has conservation of momentum, but not conservation of kinetic energy.
Momentum is conserved in any collision, but kinetic energy is only conserved in elastic collisions. So, an inelastic collision has conservation of momentum, but not conservation of kinetic energy.
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Two children are playing with sleds on a snow-covered hill. Sam weighs 50kg, and his sled weighs 10kg. Sally weighs 40kg, and her sled weighs 12kg. When they arrive, they climb up the hill using boots. Halfway up the 50-meter hill, Sally slips and rolls back down to the bottom. Sam continues climbing, and eventually Sally joins him at the top.
They then decide to sled down the hill, but disagree about who will go first.
Scenario 1:
Sam goes down the hill first, claiming that he will reach a higher velocity. If Sally had gone first, Sam says they could collide.
Scenario 2:
Sally goes down the hill first, claiming that she will experience lower friction and thus reach a higher velocity. If Sam had gone first, Sally says they could collide.
Scenario 3:
Unable to agree, Sam and Sally tether themselves with a rope and go down together.
If the hill is frictionless, what momentum does Sam experience when he reaches the bottom of the 50m hill on his sled in Scenario 1?
Two children are playing with sleds on a snow-covered hill. Sam weighs 50kg, and his sled weighs 10kg. Sally weighs 40kg, and her sled weighs 12kg. When they arrive, they climb up the hill using boots. Halfway up the 50-meter hill, Sally slips and rolls back down to the bottom. Sam continues climbing, and eventually Sally joins him at the top.
They then decide to sled down the hill, but disagree about who will go first.
Scenario 1:
Sam goes down the hill first, claiming that he will reach a higher velocity. If Sally had gone first, Sam says they could collide.
Scenario 2:
Sally goes down the hill first, claiming that she will experience lower friction and thus reach a higher velocity. If Sam had gone first, Sally says they could collide.
Scenario 3:
Unable to agree, Sam and Sally tether themselves with a rope and go down together.
If the hill is frictionless, what momentum does Sam experience when he reaches the bottom of the 50m hill on his sled in Scenario 1?
The momentum is equal to Sam's total mass times his velocity. His velocity can be found from his potential energy, using conservation of energy. Keep in mind, his total mass includes the mass of his sled.
PE = mgh = (50kg + 10kg) * 10m/s2 * 50m
PE = 30,000J
KE = 30,000J = 1/2mv2 = 1/2(60kg)(v2)
v2 = 1000 m2/s2 = 31.2m/s
p = mv = 60 kg * 31.2 m/s = 1898 kg*m/s
If you had used 30J in place of 30kJ in the KE calculation, you would have chosen 60kg*m/s
The momentum is equal to Sam's total mass times his velocity. His velocity can be found from his potential energy, using conservation of energy. Keep in mind, his total mass includes the mass of his sled.
PE = mgh = (50kg + 10kg) * 10m/s2 * 50m
PE = 30,000J
KE = 30,000J = 1/2mv2 = 1/2(60kg)(v2)
v2 = 1000 m2/s2 = 31.2m/s
p = mv = 60 kg * 31.2 m/s = 1898 kg*m/s
If you had used 30J in place of 30kJ in the KE calculation, you would have chosen 60kg*m/s
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Two children are playing with sleds on a snow-covered hill. Sam weighs 50kg, and his sled weighs 10kg. Sally weighs 40kg, and her sled weighs 12kg. When they arrive, they climb up the hill using boots. Halfway up the 50-meter hill, Sally slips and rolls back down to the bottom. Sam continues climbing, and eventually Sally joins him at the top.
They then decide to sled down the hill, but disagree about who will go first.
Scenario 1:
Sam goes down the hill first, claiming that he will reach a higher velocity. If Sally had gone first, Sam says they could collide.
Scenario 2:
Sally goes down the hill first, claiming that she will experience lower friction and thus reach a higher velocity. If Sam had gone first, Sally says they could collide.
Scenario 3:
Unable to agree, Sam and Sally tether themselves with a rope and go down together.
A third boy, John, comes to play on a neighboring hill. Sally goes to play with him, and they find that after they sled down the first hill, Sally was traveling 12m/s. John was only moving at 8m/s. John weighs 80kg with his sled. Which of the following is true?
Two children are playing with sleds on a snow-covered hill. Sam weighs 50kg, and his sled weighs 10kg. Sally weighs 40kg, and her sled weighs 12kg. When they arrive, they climb up the hill using boots. Halfway up the 50-meter hill, Sally slips and rolls back down to the bottom. Sam continues climbing, and eventually Sally joins him at the top.
They then decide to sled down the hill, but disagree about who will go first.
Scenario 1:
Sam goes down the hill first, claiming that he will reach a higher velocity. If Sally had gone first, Sam says they could collide.
Scenario 2:
Sally goes down the hill first, claiming that she will experience lower friction and thus reach a higher velocity. If Sam had gone first, Sally says they could collide.
Scenario 3:
Unable to agree, Sam and Sally tether themselves with a rope and go down together.
A third boy, John, comes to play on a neighboring hill. Sally goes to play with him, and they find that after they sled down the first hill, Sally was traveling 12m/s. John was only moving at 8m/s. John weighs 80kg with his sled. Which of the following is true?
Kinetic energy emphasizes velocity, by squaring the velocity term.
KE = 1/2mv2
Sally's KE = 1/2(52 kg)(12 m/s)2 = 3,744J
John's KE = 1/2(80 kg)(8m/s)2 = 2,560J
p = mv
Sally's Momentum = 52kg * 12m/s = 624 kg*m/s
John's Momentum = 80kg * 8m/s = 640 kg*m/s
Kinetic energy emphasizes velocity, by squaring the velocity term.
KE = 1/2mv2
Sally's KE = 1/2(52 kg)(12 m/s)2 = 3,744J
John's KE = 1/2(80 kg)(8m/s)2 = 2,560J
p = mv
Sally's Momentum = 52kg * 12m/s = 624 kg*m/s
John's Momentum = 80kg * 8m/s = 640 kg*m/s
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Two children are playing on an icy lake. Child 1 weighs 50kg, and child 2 weighs 38kg. Child 1 has a backpack that weighs 10kg, and child 2 has a backpack that weighs 5kg.
Over the course of the afternoon, they collide many times. Four collisions are described below.
Collision 1:
Child 1 starts from the top of a ramp, and after going down, reaches the lake surface while going 
and subsequently slides into a stationary child 2. They remain linked together after the collision.
Collision 2:
Child 1 and child 2 are sliding in the same direction. Child 2, moving at 
, slides into child 1, moving at 
.
Collision 3:
The two children collide while traveling in opposite directions at 
each.
Collision 4:
The two children push off from one another’s back, and begin moving in exactly opposite directions. Child 2 moves with a velocity of 
.
In all of the above scenarios, which of the following quantities is conserved?
I. Kinetic energy
II. Potential energy
III. Momentum
Two children are playing on an icy lake. Child 1 weighs 50kg, and child 2 weighs 38kg. Child 1 has a backpack that weighs 10kg, and child 2 has a backpack that weighs 5kg.
Over the course of the afternoon, they collide many times. Four collisions are described below.
Collision 1:
Child 1 starts from the top of a ramp, and after going down, reaches the lake surface while going
Collision 2:
Child 1 and child 2 are sliding in the same direction. Child 2, moving at
Collision 3:
The two children collide while traveling in opposite directions at
Collision 4:
The two children push off from one another’s back, and begin moving in exactly opposite directions. Child 2 moves with a velocity of
In all of the above scenarios, which of the following quantities is conserved?
I. Kinetic energy
II. Potential energy
III. Momentum
Kinetic energy and potential energy are interconverted. While total energy is conserved, kinetic energy is allowed to increase or decrease, provided that potential energy does the opposite. For example, as the child in collision 1 is starting at the top of a ramp, she has potential energy, but no kinetic energy. At the bottom of the ramp, all the potential energy has been turned into kinetic energy.
Momentum, on the other hand, is conserved in every collision.
Kinetic energy and potential energy are interconverted. While total energy is conserved, kinetic energy is allowed to increase or decrease, provided that potential energy does the opposite. For example, as the child in collision 1 is starting at the top of a ramp, she has potential energy, but no kinetic energy. At the bottom of the ramp, all the potential energy has been turned into kinetic energy.
Momentum, on the other hand, is conserved in every collision.
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Two children are playing on an icy lake. Child 1 weighs 50kg, and child 2 weighs 38kg. Child 1 has a backpack that weighs 10kg, and child 2 has a backpack that weighs 5kg.
Over the course of the afternoon, they collide many times. Four collisions are described below.
Collision 1:
Child 1 starts from the top of a ramp, and after going down, reaches the lake surface while going 
and subsequently slides into a stationary child 2. They remain linked together after the collision.
Collision 2:
Child 1 and child 2 are sliding in the same direction. Child 2, moving at 
, slides into child 1, moving at 
.
Collision 3:
The two children collide while traveling in opposite directions at 
each.
Collision 4:
The two children push off from one another’s back, and begin moving in exactly opposite directions. Child 2 moves with a velocity of 
.
In collision 1, with what velocity will the children be traveling after the collision?
Assume that the velocity with which child 1 hits child 2 is the same as the velocity with which child 1 reaches the surface of the lake. Also assume that the original direction of travel for child 1 is positive. Ignore friction and wind resistance.
Two children are playing on an icy lake. Child 1 weighs 50kg, and child 2 weighs 38kg. Child 1 has a backpack that weighs 10kg, and child 2 has a backpack that weighs 5kg.
Over the course of the afternoon, they collide many times. Four collisions are described below.
Collision 1:
Child 1 starts from the top of a ramp, and after going down, reaches the lake surface while going
Collision 2:
Child 1 and child 2 are sliding in the same direction. Child 2, moving at
Collision 3:
The two children collide while traveling in opposite directions at
Collision 4:
The two children push off from one another’s back, and begin moving in exactly opposite directions. Child 2 moves with a velocity of
In collision 1, with what velocity will the children be traveling after the collision?
Assume that the velocity with which child 1 hits child 2 is the same as the velocity with which child 1 reaches the surface of the lake. Also assume that the original direction of travel for child 1 is positive. Ignore friction and wind resistance.
Momentum is always conserved in collisions, and is equal to the product of mass and velocity.
Momentum is always conserved in collisions, and is equal to the product of mass and velocity.
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Two children are playing on an icy lake. Child 1 weighs 50kg, and child 2 weighs 38kg. Child 1 has a backpack that weighs 10kg, and child 2 has a backpack that weighs 5kg.
Over the course of the afternoon, they collide many times. Four collisions are described below.
Collision 1:
Child 1 starts from the top of a ramp, and after going down, reaches the lake surface while going 
and subsequently slides into a stationary child 2. They remain linked together after the collision.
Collision 2:
Child 1 and child 2 are sliding in the same direction. Child 2, moving at 
, slides into child 1, moving at 
.
Collision 3:
The two children collide while traveling in opposite directions at 
each.
Collision 4:
The two children push off from one another’s back, and begin moving in exactly opposite directions. Child 2 moves with a velocity of 
.
In collision 4, what is the final velocity of child 1? Ignore friction and air resistance.
Two children are playing on an icy lake. Child 1 weighs 50kg, and child 2 weighs 38kg. Child 1 has a backpack that weighs 10kg, and child 2 has a backpack that weighs 5kg.
Over the course of the afternoon, they collide many times. Four collisions are described below.
Collision 1:
Child 1 starts from the top of a ramp, and after going down, reaches the lake surface while going
Collision 2:
Child 1 and child 2 are sliding in the same direction. Child 2, moving at
Collision 3:
The two children collide while traveling in opposite directions at
Collision 4:
The two children push off from one another’s back, and begin moving in exactly opposite directions. Child 2 moves with a velocity of
In collision 4, what is the final velocity of child 1? Ignore friction and air resistance.
This is a reverse collision, and momentum is still conserved. The original momentum is zero.
This is a reverse collision, and momentum is still conserved. The original momentum is zero.
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Two astronauts, Ann and Bob, conduct a collision experiment in a weightless, frictionless environment. Initially Ann moves to the right with a momentum of 
, and Bob is initially at rest. In the collision, the two astronauts push on each other so that Ann's final momentum is 
to the left. What is Bob's final momentum?
Two astronauts, Ann and Bob, conduct a collision experiment in a weightless, frictionless environment. Initially Ann moves to the right with a momentum of
Apply conservation of momentum before and after the collision.
.
Taking left to be the negative direction, and noting that Bob's initial momentum is 0 since he is at rest, we can use the provided information to see that 
.
Solving for 
, we get 
. Since this answer is positive, Bob's momentum is in the positive direction (to the right).
Apply conservation of momentum before and after the collision.
Taking left to be the negative direction, and noting that Bob's initial momentum is 0 since he is at rest, we can use the provided information to see that
Solving for
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In a totally inelastic collision between two objects, all of the following are true, except that __________.
In a totally inelastic collision between two objects, all of the following are true, except that __________.
The difference between an elastic and inelastic collision is that in an elastic collision BOTH momentum and kinetic energy are conserved. Momentum is conserved in any collision, but when a collision is inelastic, kinetic energy is dissipated when the two individual objects combine. This energy can be in the form of heat, explosion, sound, etc. In a totally inelastic collision, the two objects colliding stick together and travel with the same velocity in one direction.
Overall energy is conserved, as the energy is merely transferred from a kinetic form to a non-kinetic form.
The difference between an elastic and inelastic collision is that in an elastic collision BOTH momentum and kinetic energy are conserved. Momentum is conserved in any collision, but when a collision is inelastic, kinetic energy is dissipated when the two individual objects combine. This energy can be in the form of heat, explosion, sound, etc. In a totally inelastic collision, the two objects colliding stick together and travel with the same velocity in one direction.
Overall energy is conserved, as the energy is merely transferred from a kinetic form to a non-kinetic form.
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In an elastic collision, which of the following quantities is not conserved?
In an elastic collision, which of the following quantities is not conserved?
In an _in_elastic collision, kinetic energy is not conserved. In an elastic collision, kinetic energy is conserved and there is no transfer of energy to the surroundings.
Momentum is conserved regardless of the type of collision. All of the given quantities are conserved during an elastic collision.
In an _in_elastic collision, kinetic energy is not conserved. In an elastic collision, kinetic energy is conserved and there is no transfer of energy to the surroundings.
Momentum is conserved regardless of the type of collision. All of the given quantities are conserved during an elastic collision.
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Which object has the most momentum?
I. A 
ball thrown at 
II. A 
bird diving at 
III. An 
book sliding across a desk at 
Which object has the most momentum?
I. A
II. A
III. An
We can find the momentum for each object using the momentum formula, 
.
We can find the momentum for each object using the momentum formula,
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A 
billiard ball moving at 
to the right hits a 
bowling ball at rest. The billiard ball then moves in the opposite direction at 
.
What is the velocity of the bowling ball after the collision?
A
What is the velocity of the bowling ball after the collision?
We can set this up as a conservation of momentum problem. The sum of the initial momentum for each object must equal the sum of the final momentum of each object.
We have two objects and four different velocities. The system must conserve momentum between the time before the collision and after the collision. We are given the mass of each object and their initial velocities.
We are given the magnitude of the billiard ball's final velocity, and told that it changes directions, thus becoming negative. Remember that both velocity and momentum are vector quantities.
Finally, we can solve of the final velocity of the bowling ball. We can predict that the bowling ball's velocity will be positive since the net initial velocity is positive, and the final billiard ball velocity is negative. The bowling ball must compensate for the negative momentum of the billiard ball.
We can set this up as a conservation of momentum problem. The sum of the initial momentum for each object must equal the sum of the final momentum of each object.
We have two objects and four different velocities. The system must conserve momentum between the time before the collision and after the collision. We are given the mass of each object and their initial velocities.
We are given the magnitude of the billiard ball's final velocity, and told that it changes directions, thus becoming negative. Remember that both velocity and momentum are vector quantities.
Finally, we can solve of the final velocity of the bowling ball. We can predict that the bowling ball's velocity will be positive since the net initial velocity is positive, and the final billiard ball velocity is negative. The bowling ball must compensate for the negative momentum of the billiard ball.
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