Year 11 Physics Momentum, Energy & Simple Systems Practice Test (Free)

Question 1

A

1500\text{ kg}

car travels at

20\text{ m/s}

. What is its momentum?

A

75\text{ kg·m/s}

B

3000\text{ kg·m/s}

C

3.0\times10^5\text{ kg·m/s}

D

3.0\times10^4\text{ kg·m/s}

E

1520\text{ kg·m/s}

Show worked solution

Correct: D

$p = mv = (1500)(20) = 3.0\times10^4\text{ kg·m/s}$ .

Question 2

A ball has a momentum of

12\text{ kg·m/s}

and a mass of

0.50\text{ kg}

. How fast is it moving?

A

48\text{ m/s}

B

12.5\text{ m/s}

C

24\text{ m/s}

D

6.0\text{ m/s}

E

0.042\text{ m/s}

Show worked solution

Correct: C

Rearrange $p = mv$ : $v = \dfrac{p}{m} = \dfrac{12}{0.50} = 24\text{ m/s}$ .

Question 3

A constant

50\text{ N}

force acts on a ball for

0.20\text{ s}

. What impulse does it deliver?

A

250\text{ N·s}

B

2.5\text{ N·s}

C

50\text{ N·s}

D

0.004\text{ N·s}

E

10\text{ N·s}

Show worked solution

Correct: E

$J = Ft = (50)(0.20) = 10\text{ N·s}$ .

Question 4

A

2.0\text{ kg}

object's velocity changes from

3.0\text{ m/s}

to

8.0\text{ m/s}

. What is its change in momentum?

A

5.0\text{ kg·m/s}

B

10\text{ kg·m/s}

C

16\text{ kg·m/s}

D

2.5\text{ kg·m/s}

E

22\text{ kg·m/s}

Show worked solution

Correct: B

$\Delta p = m\Delta v = m(v - u) = 2.0(8.0 - 3.0) = 10\text{ kg·m/s}$ .

Question 5

The impulse acting on an object is always equal to its...

Show worked solution

Correct: E

The impulse-momentum theorem states $J = Ft = \Delta p$ : the impulse equals the change in momentum. This is just Newton's second law written for a time interval.

Question 6

Two objects collide with no external forces acting. Compared with before the collision, the total momentum afterward is...

Show worked solution

Correct: C

In an isolated system (no external forces) the total momentum is conserved, it is the same before and after any interaction. This holds for every collision and explosion.

Question 7

The graph shows the constant force acting on a ball during a

0.5\text{ s}

kick. What impulse does the force deliver?

A

0.5\text{ N·s}

B

200\text{ N·s}

C

40\text{ N·s}

D

100\text{ N·s}

E

400\text{ N·s}

Show worked solution

Correct: D

The impulse is the area under a force-time graph: $J = (200)(0.5) = 100\text{ N·s}$ .

Question 8

A

0.15\text{ kg}

ball hits a wall at

20\text{ m/s}

and rebounds straight back at

20\text{ m/s}

. If the contact lasts

0.10\text{ s}

, what is the average force from the wall?

Show worked solution

Correct: D

Taking "in" as positive, the velocity changes from $+20$ to $-20\text{ m/s}$ , a change of $40\text{ m/s}$ : $\Delta p = m\Delta v = 0.15(40) = 6.0\text{ kg·m/s}$ . Then $F = \dfrac{\Delta p}{t} = \dfrac{6.0}{0.10} = 60\text{ N}$ . (Using only $20$ gives the $30\text{ N}$ trap; a bounce reverses the velocity, so the change is $40$ , not $0$ .)

Question 9

A

2.0\text{ kg}

cart moving at

3.0\text{ m/s}

collides with a stationary

1.0\text{ kg}

cart and they lock together. What is their common velocity afterward?

A

6.0\text{ m/s}

B

1.5\text{ m/s}

C

3.0\text{ m/s}

D

1.0\text{ m/s}

E

2.0\text{ m/s}

Show worked solution

Correct: E

Momentum is conserved: $m_1u_1 = (m_1 + m_2)v$ , so $(2.0)(3.0) = (3.0)v$ and $v = \dfrac{6.0}{3.0} = 2.0\text{ m/s}$ . (The combined mass must be used, not just the first cart's.)

Question 10

A

60\text{ kg}

person on a frictionless skateboard, initially at rest, throws a

4.0\text{ kg}

ball horizontally at

5.0\text{ m/s}

. What is the person's recoil speed?

A

5.0\text{ m/s}

B

0.33\text{ m/s}

C

20\text{ m/s}

D

0.083\text{ m/s}

E

3.0\text{ m/s}

Show worked solution

Correct: B

Total momentum starts at zero: $0 = (4.0)(5.0) + (60)v$ , so $v = -\dfrac{20}{60} \approx -0.33\text{ m/s}$ , i.e. $0.33\text{ m/s}$ in the opposite direction to the ball.

Question 11

A slow, heavy truck and a fast, light bullet happen to have exactly the same momentum. Which one has the greater kinetic energy?

AThe bullet — for equal momentum, the smaller mass has the greater kinetic energyBIt cannot be found without knowing the speedsCThe truck — it has the greater massDNeither — equal momentum means zero kinetic energyEThey are equal, because their momenta are equal

Show worked solution

Correct: A

Kinetic energy is linked to momentum by $KE = \dfrac{p^2}{2m}$ . For the same momentum $p$ , the smaller mass gives the larger kinetic energy, so the bullet has far more. (Equal momentum does not mean equal kinetic energy, which is why a small, fast bullet does so much damage.)

Question 12

The force on a ball rises and falls as shown, reaching a peak of

80\text{ N}

over a total time of

0.5\text{ s}

. What impulse does it deliver?

A

40\text{ N·s}

B

20\text{ N·s}

C

10\text{ N·s}

D

160\text{ N·s}

E

80\text{ N·s}

Show worked solution

Correct: B

The impulse is the area of the triangle: $J = \tfrac{1}{2}(\text{base})(\text{height}) = \tfrac{1}{2}(0.5)(80) = 20\text{ N·s}$ . (Forgetting the $\tfrac{1}{2}$ gives the $40\text{ N·s}$ trap.)

Question 13

Modern cars have airbags and crumple zones. How do they reduce the force on the occupants in a crash?

AThey increase the car's momentumBThey add an equal and opposite forceCThey reduce the change in momentumDThey reduce the occupant's massEThey lengthen the collision time, so the same change in momentum needs a smaller force

Show worked solution

Correct: E

The change in momentum (from full speed to rest) is fixed. Since $Ft = \Delta p$ , stretching out the time $t$ makes the force $F$ smaller. Airbags and crumple zones extend the stopping time.

Question 14

In a perfectly inelastic collision (the objects stick together), what is conserved?

Show worked solution

Correct: A

Momentum is conserved in every collision, but kinetic energy is not conserved in an inelastic collision, some is converted to heat, sound and deformation. (In an elastic collision, both are conserved.)

Question 15

A

3.0\text{ kg}

rifle fires a

0.010\text{ kg}

bullet at

400\text{ m/s}

. What is the recoil speed of the rifle?

A

4.0\text{ m/s}

B

0.40\text{ m/s}

C

40\text{ m/s}

D

1.3\text{ m/s}

E

0.013\text{ m/s}

Show worked solution

Correct: D

Total momentum starts at zero: $0 = (0.010)(400) + (3.0)v$ , so $v = -\dfrac{4.0}{3.0} \approx -1.3\text{ m/s}$ . The rifle recoils at $1.3\text{ m/s}$ backward.

Question 16

A

2.0\text{ kg}

cart moving at

5.0\text{ m/s}

strikes a stationary

3.0\text{ kg}

cart. Afterward the

2.0\text{ kg}

cart continues forward at

1.0\text{ m/s}

. What is the velocity of the

3.0\text{ kg}

cart?

A

2.67\text{ m/s}

B

3.33\text{ m/s}

C

1.67\text{ m/s}

D

5.0\text{ m/s}

E

8.0\text{ m/s}

Show worked solution

Correct: A

Conserve momentum: $(2.0)(5.0) = (2.0)(1.0) + (3.0)v$ , so $10 = 2 + 3v$ , giving $v = \dfrac{8}{3} \approx 2.67\text{ m/s}$ .

Question 17

A

2.0\text{ kg}

trolley at

3.0\text{ m/s}

sticks to a stationary

1.0\text{ kg}

trolley, and the pair moves off at

2.0\text{ m/s}

. How much kinetic energy is lost in the collision?

Show worked solution

Correct: A

Before: $KE = \tfrac{1}{2}(2.0)(3.0)^2 = 9.0\text{ J}$ . After: $KE = \tfrac{1}{2}(3.0)(2.0)^2 = 6.0\text{ J}$ . Lost: $9.0 - 6.0 = 3.0\text{ J}$ , converted to heat and sound. (Momentum is conserved, but kinetic energy is not.)

Question 18

A

0.020\text{ kg}

bullet travelling at

300\text{ m/s}

embeds itself in a stationary

2.0\text{ kg}

wooden block. What is their common velocity just after impact?

A

150\text{ m/s}

B

0.30\text{ m/s}

C

6.0\text{ m/s}

D

3.0\text{ m/s}

E

2.97\text{ m/s}

Show worked solution

Correct: E

Momentum is conserved: $(0.020)(300) = (0.020 + 2.0)v$ , so $v = \dfrac{6.0}{2.02} \approx 2.97\text{ m/s}$ . (Ignoring the bullet's mass in the total gives $6.0/2.0 = 3.0$ , the tempting trap.)

Question 19

The force-time graph shown acts on a

5.0\text{ kg}

object initially at rest, delivering an impulse of

30\text{ N·s}

(the shaded area). What is the object's final velocity?

A

6.0\text{ m/s}

B

30\text{ m/s}

C

0.17\text{ m/s}

D

3.0\text{ m/s}

E

150\text{ m/s}

Show worked solution

Correct: A

The impulse equals the change in momentum: $J = m\Delta v$ , so $\Delta v = \dfrac{J}{m} = \dfrac{30}{5.0} = 6.0\text{ m/s}$ . Starting from rest, the final velocity is $6.0\text{ m/s}$ .

Question 20

A

4.0\text{ kg}

object at rest suddenly explodes into two pieces. A

1.0\text{ kg}

piece flies off at

6.0\text{ m/s}

. What is the velocity of the

3.0\text{ kg}

piece?

A2.0 m/s in the opposite directionB2.0 m/s in the same directionC18 m/s in the opposite directionD6.0 m/s in the opposite directionE0.5 m/s in the same direction

Show worked solution

Correct: A

Total momentum stays zero: $0 = (1.0)(6.0) + (3.0)v$ , so $v = -\dfrac{6.0}{3.0} = -2.0\text{ m/s}$ , i.e. $2.0\text{ m/s}$ in the opposite direction. The lighter piece moves faster.

Question 21

A moving ball makes a head-on elastic collision with a second, identical ball that is at rest. What happens?

AThe first ball bounces straight backBThe first ball stops and the second moves off at the first ball's speedCBoth balls stopDBoth balls move off at half the speedEThey stick together and move off at half the speed

Show worked solution

Correct: B

For an elastic collision between equal masses, the two velocities are exchanged. The incoming ball stops dead and the target ball moves off with the incoming ball's velocity, conserving both momentum and kinetic energy (as seen in Newton's cradle).

Question 22

A

70\text{ kg}

driver moving at

15\text{ m/s}

is brought to rest in a crash. With an airbag the stop takes

0.50\text{ s}

. What average force acts on the driver?

Show worked solution

Correct: E

$F = \dfrac{\Delta p}{t} = \dfrac{m\Delta v}{t} = \dfrac{(70)(15)}{0.50} = 2100\text{ N}$ . Without the airbag the stop might take only $0.05\text{ s}$ , giving ten times the force ( $21{,}000\text{ N}$ ), which is why airbags matter.

Question 23

A

3.0\text{ kg}

trolley moving east at

2.0\text{ m/s}

collides head-on with a

2.0\text{ kg}

trolley moving west at

5.0\text{ m/s}

, and they stick together. What is their common velocity?

Show worked solution

Correct: D

Take east as positive: $(3.0)(+2.0) + (2.0)(-5.0) = 6.0 - 10 = -4.0\text{ kg·m/s}$ . Then $v = \dfrac{-4.0}{5.0} = -0.80\text{ m/s}$ , i.e. $0.80\text{ m/s west}$ (the faster, westward trolley wins).

Question 24

Two balls collide. Before: a

2.0\text{ kg}

ball moves at

6.0\text{ m/s}

toward a stationary

4.0\text{ kg}

ball. After: the

2.0\text{ kg}

ball moves at

-2.0\text{ m/s}

and the

4.0\text{ kg}

ball at

4.0\text{ m/s}

. Is the collision elastic?

AYes — because momentum is conservedBCannot tell without more informationCNo — kinetic energy is not conservedDNo — momentum is not conservedEYes — kinetic energy is conserved (36 J before and after)

Show worked solution

Correct: E

Momentum checks out either way, so test the kinetic energy. Before: $\tfrac{1}{2}(2)(6)^2 = 36\text{ J}$ . After: $\tfrac{1}{2}(2)(2)^2 + \tfrac{1}{2}(4)(4)^2 = 4 + 32 = 36\text{ J}$ . Kinetic energy is conserved, so it is elastic. (Momentum is conserved in every collision, so that alone never proves it is elastic.)

Question 25

A

1000\text{ kg}

car moving at

20\text{ m/s}

runs into the back of a stationary

1500\text{ kg}

car and they lock together. What fraction of the original kinetic energy is lost in the collision?

Show worked solution

Correct: E

Common velocity: $v = \dfrac{(1000)(20)}{2500} = 8.0\text{ m/s}$ . Before: $KE = \tfrac{1}{2}(1000)(20)^2 = 2.0\times10^5\text{ J}$ . After: $KE = \tfrac{1}{2}(2500)(8.0)^2 = 8.0\times10^4\text{ J}$ . Lost: $1.2\times10^5\text{ J}$ , which is $\dfrac{1.2\times10^5}{2.0\times10^5} = 60\%$ .