MCAT Physical : MCAT Physical Sciences
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Example Questions
Example Question #31 : Translational Motion
A car is moving with a constant velocity of 
This problem can be easily solved using the formula, 
The car comes to a stop after applying the breaks over 10m.
Example Question #262 : Mcat Physical Sciences
A sprinter running a 
To answer this question, we must have a solid understanding of the kinematics equations. For this question, we must use an equation relating final velocity, distance, and acceleration.
The best fit for this is 
Since we are solving for final velocity, and we started from rest, we can simplify the equation.
Example Question #261 : Mcat Physical Sciences
An car accelerates from rest and travels 
The kinematics equation relating distance, time, and acceleration is 
Since we know that the car was initially at rest, we can rewrite the equation with zero initial velocity.
Now we can plug in our values from the question, and solve for the acceleration.
Example Question #261 : Mcat Physical Sciences
If an ball is dropped from a cliff 
Since acceleration is constant, we can use the appropriate kinematics equation to solve:
We are given the height of the cliff, which will be equal to the distance traveled. The initial velocity is zero since the ball starts from rest and the acceleration will be equal to the acceleration due to gravity. Use these values to calculate the time.
Example Question #11 : Motion In One Dimension
A 2kg mass is suspended on a rope that wraps around a frictionless pulley. The pulley is attached to the ceiling and has a mass of 0.01kg and a radius of 0.25m. The other end of the rope is attached to a massless suspended platform, upon which 0.5kg weights may be placed. While the system is initially at equilibrium, the rope is later cut above the weight, and the platform subsequently raised by pulling on the rope.
What is the velocity of the platform two seconds after the rope is cut?
Think back to the 3 main kinematics equations that we know.
vf2 = vi2 + 2aΔx
vf = vi + at
Δx = vit + ½at2
We need to determine which formula will allow us to find final velocity after a given amount of time.
vf = vi + at = (0m/s) + (9.8m/s2)(2s) = 19.6m/s
Example Question #262 : Mcat Physical Sciences
If an object is dropped from a height of 450 meters above Earth, what is its velocity just before impact?
67m/s
6.7m/s
95m/s
45m/s
95m/s
First calculate the time it takes to hit the ground using the equation 
We can plug in values (including acceleration due to gravity) and solve for t.
t = 9.49s
Next, find the final velocity with the equation 
Example Question #41 : Translational Motion
A sports car has a mass of 
This question will require us to deal with work and force.
We are given the mass of the car, the work energy generated, and the distance. Using these values, we can solve for the acceleration.
Example Question #22 : Motion In One Dimension
Which of the following is not a vector quantity?
Momentum
Kinetic energy
Weight
Velocity
Displacement
Kinetic energy
Kinetic energy is calculated by squaring velocity (which we know is a vector). This eliminates its vector properties making kinetic energy a scalar value. All other answer choices are vector quantities.
Example Question #1 : Motion In Two Dimensions
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?
-9.8m/s2
0m/s2
9.8m/s2
-4.9m/s2
0m/s2
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.
Example Question #92 : Newtonian Mechanics And Motion
A ball of mass 3kg is thrown into the air at an angle of 45o above the horizontal, with initial velocity of 15m/s. Instantaneously at the highest point in its motion, the ball comes to rest. Approximately what is the magnitude of acceleration at this point? Assume that air resistance is negligible.
15m/s2
5m/s2
10m/s2
0m/s2
30m/s2
10m/s2
After being thrown, the ball is only acted on by the force of gravity. Since this force is constant throughout the motion, acceleration must also remain constant, and be equal to the gravitational acceleration of 9.8 m/s2 (approximately 10 m/s2)
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