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Momentum - Multiple choice solutions

Multiple choice questions should take you just under 2 minutes a question!

Only one of the choices is correct.

 

Q1. Which one of the following statements is correct?

The force acting on an object is equivalent to:

A its change of momentum.
B the impulse it receives per second.
C the energy it gains per second.
D its acceleration per metre.

 

Ft (impulse) = p (change in momentum)

therefore F = p/t

F = Ft/t = impulse/t

Q2. The graph shows how the force on a glider of mass 2000 kg changes with time as it is launched from a level track using a catapult.

Assuming the glider starts at rest what is its velocity in ms-1 after 40 s?

A
2.5
B
10
C
50
D
100

 

Area under graph is impulse so Ft = 0.5 x 5 x 40 = 100 kNs

That is the change in momentum - initial momentum being zero we know that final momentum = 100 kNs

mv = 100,000 N

m = 2000 kg therefore v = 100,000/2000 = 50 m/s

 

Q3. A gas molecule of mass m in a container moves with velocity v.

If it makes an elastic collision at right angles to the walls of the container, what is the change in momentum of the molecule?

A
zero
B
0.5 mv
C
mv
D
2 mv

 

mv - (-mv) = 2mv

Q4. The graph shows the variation with time, t, of the force, F, acting on a body.

What physical quantity does the area X represent?

A
 the displacement of the body
B
 the acceleration of the body
C
 the change in momentum of the body
D
 the change in kinetic energy of the body

 

Q5. Water of density 1000 kg m–3 flows out of a garden hose of cross-sectional area 7.2 × 10–4m2 at a rate of 2.0 × 10–4m3 per second.

How much momentum is carried by the water leaving the hose per second?

A
 5.6 × 10–5 Ns
B
 5.6 × 10–2 Ns
C
 0.20 Ns
D
 0.72 Ns

 

density = mass/volume

therefore: mass = density x volume.

You can then work out the mass delivered every second.

Volume delivered every second = 2.0 × 10–4m3

Therefore mass delivered every second = 1000 x 2.0 × 10–4 kg

Therefore mass delivered every second = 0.2 kg

The water is delivered like a 'tube' - you are given the CSA so you can work out the length of that tube. The length of that tube is the distance the water travels in a second - the velocity!

volume/s = csa x length = 2.0 × 10–4m3

velocity = 2.0 × 10–4 / 7.2 × 10–4 = 0.278 m/s

Momentum = mv

Momentum = 0.2 x 0.278 = 0.0556 Ns

Choice B

Q6. Which row, A to D, in the table correctly shows the quantities conserved in an inelastic
collision?

 
mass
momentum
kinetic energy
total energy
A conserved not conserved conserved conserved
B not conserved conserved conserved not conserved
C conserved conserved conserved conserved
D conserved conserved not conserved conserved

 

Inelastic collision - KE not conserved - a really quick one to answer!

Q7. A ball of mass 2.0 kg, initially at rest, is acted on by a force F which varies with time t as shown by the graph. What is the velocity of the ball after 8.0 s?

Impulse = area under graph = 40 Ns

The impulse produces the change in momemtum, which, as the initial velocity was zero, will be equal to the final momemtum after 8 seconds.

Final mv = 40, but m = 2.0 kg so v = 20 m/s

A
20 ms–1
B
40 ms–1
C
80 ms–1
D
160 ms–1

 

Q8. A body X moving with a velocity v makes an elastic collision with a stationary body Y of equal mass on a smooth horizontal surface.

Which line, A to D, in the table gives the velocities of the two bodies after the collision?
Velocity of X
Velocity of Y
A
v/2
- v/2
B
- v/2
v/2
C
v
0
D
0
v

 

Elastic collision - therefore no loss of KE.

Initial KE = ½mv2 but if velocities halve total KE will be halved therefore A and B cannot happen.

The ball cannot continue at velocity v without pushing the other ball too - therefore C is wrong.

Answer must be D

 

Q9. Which of the following is a scalar quantity?

A
momentum
B
weight
C
power
D
moment

 

Simples - if you know your scalar/vector quantities!

 

Q10. A car of mass 580 kg collides with the rear of a stationary van of mass 1200 kg.

Following the collision, the van moves with a velocity of 6.20 m s−1 and the car recoils in the opposite direction with a velocity of 1.60 m s−1 .

What is the initial speed of the car?

A
 5.43 m s−1
B
 11.2 m s−1
C
 12.8 m s−1
D
14.4 m s−1

 

Momentum (mv) is conserved

total momentum before collision = total momentum after the collision

580 x v = 1200 x 6.2 - 580 x 1.6

v = (7440 - 928)/580

v = 11.2 m s−1

Choice B

 

Q11. A bullet of mass 10 g is fired with a velocity of 100 ms–1 from a stationary rifle of mass 4.0 kg.

Consider the rifle and bullet to be an isolated system.

Calculate the recoil velocity of the rifle and the total momentum of the rifle and bullet just after firing.
Recoil velocity / m s–1
Total momentum / kg m s–1
A
0.25
0
B
0.25
1.0
C
0.40
0
D
0.40
1.0

 

Momentum before firing = mbulletVbullet + MrifleVrifle

Momentum before firing = 0.010 x 0 + 4.0 x 0 = zero

Momentum after firing = mbulletVbullet + MrifleVrifle

Momentum after firing = 0.010 x 100 + 4.0 x Vrifle

By the conservation of momentum:

Momentum before firing = Momentum after firing = 0

0 = 1 + 4Vrifle

Vrifle= - 0.25 m/s

Choice A

Q12. Which row correctly states whether momentum, mass and velocity are scalar or vector quantities?
Momentum
Mass
Velocity
A
scalar
scalar
vector
B
vector
scalar
scalar
C
scalar
vector
scalar
D
vector
scalar
vector

 

Simples - if you know your scalar/vector quantities!

Choice D

Q13.The graph shows how the force F applied to an object varies with time t.

What is the momentum gained by the object from t = 0 to t = 10 s?

Ft = Δp

This is the area under the graph

Area under the graph between 0 s and 10 s = (4 x 10) + ½(6 x 6)

= 40 + 18 = 58 Ns

Choice D