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1. Define momentum.
Momentum is defined as being the product of the mass of an object and it’s velocity, or the ‘quantity of motion’.
2. Is momentum a vector or a scalar quantity? Answer
You must provide a direction as well in your final answer.
3. A trolley X of mass 6kg has a velocity of 4ms-1 east.
Calculate it’s momentum.
p = mv
p = 6(4)
p = 24kg.m.s-1 east
4. Calculate the change in momentum of a 3kg trolley if the velocity of the trolley increased from 5ms-1 east to 7ms-1 east. Answer
Δp = m(vf - vi)
Δp = 3(7 - 5)
Δp = 6kg.m.s-1 east
5. Calculate the change in momentum of a 5kg trolley if the velocity of the trolley changed from 8ms-1 east to 12ms-1 west. Answer
Choose final direction as positive
Thus west is positive, and east is negative.

Δp = m(vf - vi)
Δp = 5(12 - (-8))
Δp = 5(20)
Δp = 100 kg.m.s-1
Since answer is positive,
Δp is 100 kg.m.s-1 west

6. A force of 4N acts west on a trolley for 3s. Calculate the impulse exerted on the trolley. Answer
Δp is also called IMPULSE
Δp can also be calculated from:

Δp = FΔt
Δp = 4(3)
Δp = 12N.s west

7. The velocity of a 2kg trolley changed from 4ms-1 east to 7ms-1 east over a period of 10s. Calculate the force that must have been used. Answer
Since Δp = m(vf - vi)
and Δp = FΔt

FΔt = m(vf - vi)
F(10) = 2(7 - 4)
F = 0,6 N east.

8. An arrow of mass 250g strikes a wooden block with a velocity of 80ms-1 east and is brought to a halt in 0,8s. Calculate the stopping force. Answer
m = 0,25kg
vf = 0 (stopped) positive = east

FΔt = m(vf - vi)
F(0,8) = 0,25(0 - 80)
F(0,8) = -20
F = -25N
F = 25N west
(the negative sign means west)

9. The rate of change of momentum is equivalent to :

A. impulse
B. force
C. energy
D. power

B. force

FΔt = Δp
F = Δp/Δt
The right hand side is called "rate of change of momentum" and it is equal to force (resultant).

10. A body of mass m strikes a wall perpendicularly with a speed of v and bounces directly back with no change in its speed. The change in its momentum would be:

A. zero
B. mv
C. 2mv
D. 2v

C. 2mv
two directions = two signs
Let final velocity = v
Then initial velocity = -v

Δp = m(vf - vi)
Δp = m(v -(-v))
Δp = m(v + v)
Δp = m(2v)
Δp = 2mv

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