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MCAT Physical : Momentum

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Example Questions

Example Question #11 : Newtonian Mechanics And Motion

A $\small 1000kg$ car moving east at $\small 10\frac{m}{s}$ strikes the rear of a $\small 750kg$ car, also traveling east at $\small 5\frac{m}{s}$ , and the two stay in contact after the collision. What is the final velocity of the cars?

Possible Answers:

$5.69\frac{m}{s}$

$7.86\frac{m}{s}$

$7.5\frac{m}{s}$

$3.24\frac{m}s{}$

$12.51\frac{m}{s}$

Correct answer:

$7.86\frac{m}{s}$

Explanation:

This is a case of inelastic collision because the objects stay in contact after the impact. The total momentum in a system is always conserved, regardless of collision type. Here, the momentum before the collision is given by the sum of the individual momentum of each car, and the momentum after the collision is given by the singular momentum of the joined vehicles.

m_1v_1+m_2v_2=(m_1+m_2)v_3

We are given the mass of each car and their initial velocities, allowing us to completely solve the left side of this equation.

$(1000kg)(10\frac{m}{s})+(750kg)(5\frac{m}{s})=(m_1+m_2)v_3$

$13750\frac{kg*m}{s}=(m_1+m_2)v_3$

We can now use the combined mass of both cars to solve for their final velocity.

$13750\frac{kg*m}{s}=(1000kg+750kg)v_3$

$v_3=\frac{13750\frac{kg*m}{s}}{1750kg}=7.86\frac{m}{s}$

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Example Question #12 : Newtonian Mechanics And Motion

A $\small 100g$ ball moving at $\small 2\frac{m}{s}$ strikes a stationary $\small 50g$ solid ball head on. The smaller ball moves away at $\small 1\frac{m}{s}$ . What is the velocity of the first ball after the impact?

Possible Answers:

$0.333\frac{m}{s}$

$0.75\frac{m}{s}$

$1.5\frac{m}{s}$

$3\frac{m}{s}$

$1.15\frac{m}{s}$

Correct answer:

$1.5\frac{m}{s}$

Explanation:

Momentum is conserved in all situations. Here, we have an elastic collision, meaning that the objects do not stick together after the impact. Momentum is simply the product of mass and velocity, and it has the units $\small \frac{kg*m}{s}$ .

p=mv

We can solve this problem by setting the initial and final momentum of the system equal to each other. The momentum of the system will be the sum of the momentum of the parts.

$m_1v_{i1}+m_2v_{i2}=m_1v_{f1}+m_2v_{f2}$

The second ball contributes no momentum initially, because its initial velocity is zero. The total initial momentum comes from the initial velocity of the larger ball.

$(0.1kg)(2\frac{m}{s})+(0.05kg)(0\frac{m}{s})=m_1v_{f1}+m_2v_{f2}$

$0.2\frac{kg*m}{s}=m_1v_{f1}+m_2v_{f2}$

We know the mass of each ball and the final velocity of the smaller ball. Using these values, we can solve for the final velocity of the larger ball.

$0.2\frac{kg*m}{s}=(0.1kg)v_{f1}+(0.05kg)(1\frac{m}{s})$

$0.2\frac{kg*m}{s}=(0.1kg)v_{f1}+0.05\frac{kg*m}{s}$

$0.15\frac{kg*m}{s}=(0.1kg)v_{f1}$

$v_{f1}=\frac{0.15\frac{kg*m}{s}}{0.1kg}=1.5\frac{m}{s}$

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Example Question #11 : Translational Motion

An object of mass 3kg is moving at a velocity of $5\frac{m}{s}$ when it undergoes a perfect inelastic collision with a stationary mass. After the collision, the objects are moving at velocity of $3\frac{m}{s}$ . What is the mass of the stationary object?

Possible Answers:

6kg

1kg

2kg

4kg

Correct answer:

2kg

Explanation:

Momentum is given by the equation p=mv . Conservation of momentum states that m_1v_1=m_2v_2 .

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.

$m_1v_1=(3kg)(5\frac{m}{s})+(x)(0\frac{m}{s})=15\frac{kg*m}{s}$

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.

$m_2v_2=(3kg+xkg)(3\frac{m}{s})$

Since we know that m_1v_1=m_2v_2 , we can use our values from these calculations to solve for .

$(3kg+xkg)(3\frac{m}{s})=15\frac{kg*m}{s}$

(3kg+xkg)=5kg

x=2kg

The mass of the second object must be 2kg.

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Example Question #1 : Momentum

Ball A, traveling $7\frac{m}{s}$ to the right, collides with ball B, traveling $8\frac{m}{s}$ 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?

Possible Answers:

$5\frac{m}{s}\ \text{to the right}$

$5\frac{m}{s}\ \text{to the left}$

$6\frac{m}{s}\ \text{to the right}$

$2\frac{m}{s}\ \text{to the left}$

$2\frac{m}{s}\ \text{to the right}$

Correct answer:

$2\frac{m}{s}\ \text{to the left}$

Explanation:

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 p=mv , 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)(v_f). We can solve for v_f through conservation of momentum; the sum of the initial momentum values must equal the final momentum.

$(4kg)(7m/s) + (6kg)(-8m/s) = (10kg)(v_{f})$

Note: ball B's velocity is negative because they are traveling in opposite directions.

28 -48=10v_f

-20=10v_f

v_f = -2m/s

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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Example Question #13 : Newtonian Mechanics And Motion

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 $5\frac{m}{s}$ 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 $10\frac{m}{s}$ , slides into child 1, moving at $2\frac{m}{s}$ .

Collision 3:

The two children collide while traveling in opposite directions at $10\frac{m}{s}$ 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 $+8\frac{m}{s}$ .

In all of the above scenarios, which of the following quantities is conserved?

I. Kinetic energy

II. Potential energy

III. Momentum

Possible Answers:

I, II, and III

I and II

III only

I and III

II and III

Correct answer:

III only

Explanation:

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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Example Question #11 : 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:

Collision 2:

Child 1 and child 2 are sliding in the same direction. Child 2, moving at $10\frac{m}{s}$ , slides into child 1, moving at $2\frac{m}{s}$ .

Collision 3:

The two children collide while traveling in opposite directions at $10\frac{m}{s}$ 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 $+8\frac{m}{s}$ .

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.

Possible Answers:

$0\frac{m}{s}$

$4.1\frac{m}{s}$

$2.9\frac{m}{s}$

$-4.1\frac{m}{s}$

$-2.9\frac{m}{s}$

Correct answer:

$2.9\frac{m}{s}$

Explanation:

Momentum is always conserved in collisions, and is equal to the product of mass and velocity.

mv(before) = mv(after)

60 kg*(5 m/s) + 43 kg*(0 m/s) = (60 kg + 43 kg)(v_f)

60 kg * (5 m/s) = 103 kg * (v_f)

v_f = +2.9 m/s

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Example Question #61 : Mcat Physical Sciences

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:

Collision 2:

Child 1 and child 2 are sliding in the same direction. Child 2, moving at $10\frac{m}{s}$ , slides into child 1, moving at $2\frac{m}{s}$ .

Collision 3:

The two children collide while traveling in opposite directions at $10\frac{m}{s}$ 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 $+8\frac{m}{s}$ .

In collision 4, what is the final velocity of child 1? Ignore friction and air resistance.

Possible Answers:

$8\frac{m}{s}$

$5.7\frac{m}{s}$

$-5.7\frac{m}{s}$

Cannot be determined from the information given

$-8\frac{m}{s}$

Correct answer:

$-5.7\frac{m}{s}$

Explanation:

This is a reverse collision, and momentum is still conserved. The original momentum is zero.

0 = m_1v_1 + m_2v_2 = 43 kg * 8 m/s + 60kg * v_2

v_2 = -5.7 m/s

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Example Question #11 : Newtonian Mechanics And Motion

Which of the following is a correct description of an inelastic collision?

Possible Answers:

Neither kinetic energy nor momentum is conserved

Kinetic energy is conserved, and momentum may be conserved

Momentum is conserved, but kinetic energy is not

Both kinetic energy and momentum are conserved

Kinetic energy is conserved, but momentum is not

Correct answer:

Momentum is conserved, but kinetic energy is not

Explanation:

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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Example Question #11 : Momentum

In an elastic collision, which of the following quantities is not conserved?

Possible Answers:

Momentum

Mechanical energy

Kinetic energy

All of these are conserved

Potential energy

Correct answer:

All of these are conserved

Explanation:

In an inelastic 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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MCAT Physical : Momentum

Study concepts, example questions & explanations for MCAT Physical

All MCAT Physical Resources

Example Questions

Example Question #11 : Newtonian Mechanics And Motion

Example Question #12 : Newtonian Mechanics And Motion

Example Question #11 : Translational Motion

Example Question #1 : Momentum

Example Question #13 : Newtonian Mechanics And Motion

Example Question #11 : Momentum

Example Question #61 : Mcat Physical Sciences

Example Question #11 : Newtonian Mechanics And Motion

Example Question #11 : Momentum

All MCAT Physical Resources

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