2011 Checkpoints Chapter 4 Momentum

Question 127
C, D, E
In every collision, momentum is conserved. So the initial momentum must equal the final momentum.
Momentum (p) = mv.  pi = m × 3 + m × -6 pi = -3m
So pf = - 3m A = -7m + 4m = -3m B = 4m -m = 3m C = -3m + 0 = -3m D = -1.5 m - 1.5 m = -3m E = -5.5m + 2.5 m = -3m
At this stage A, C, D, E are all possibilities. In any collision the energy (KE + PE) must be conserved or lost to other forms. If you consider the KEi = × m × 32 + × m ×62 = × m × 45 = 22.5m
The final KE must be no greater than this. In 'A' KE = × m × 72 + × m × 42 = 32.5m.
This is greater than 22.5m.  A is impossible. C,D,E remain possible

Question 128
Momentum is conserved in all collisions.
 pi = 500  5 = 2500Ns.  pf = 2500 = 3000  1 + 500v  v = -1m/s
The answer is actually 1m/s. The negative sign means that the car travels backwards.

Question 129
Momentum is conserved if pi = pf pi = m × 2 + m × 0 = 2m (Ns) pf = m × 0.5 + m × 1.5 = 2m (Ns)
 momentum is conserved

Question 130
This is a question on conservation of momentum. The initial momentum equals the final momentum.
Pi = m1u1 + m2u2 = 55  5 + 45  1 =275 + 45 = 320
Pf = v(m1 + m2) = v  100 v = 320  100 = 3.2m/s

Question 131
C
The momentum is conserved in all collisions. KEi = ½  55  52 + ½  45  12 687.5 + 22.5 = 710 J KEf = ½  100  3.22 = 512 J
 Kinetic Energy is lost in this collision. Question 132
Malachi and Sarah will continue at the same speed and direction as they were before they let go. This is because neither have had a net force acting on them ( no change in momentum to either)

Question 133
2m/s
In every collision you will need to consider momentum will always be conserved. So as soon as you see a collision problem, you immediately think momentum, this will give you all the velocities and the masses, you can then use these to see what happens to the energies in the collision.
 pi = mv= 150 × 6 + 150 × 0 = 900 kgm/s
 pf = mv= 450 × vf  vf = 900  450 = 2 m/s.
Think about this answer. Does it make sense that when the car collides with another two the same size, and they all stick together, then the final speed will be a third of the initial speed.

Question 134
600 kgm/s
The change in momentum is given by pf - pi = 150 × 2 - 150 × 6 = 600 kgm/s.

Question 135
600 kgm/s
Impulse = change in momentum = pf - pi = 300 × 2 - 300 × 0 = 600 kgm/s
This answer must be the same as question 112. because the momentum lost by the first car must be the same as the impulse given to the other 2.

Question 136
Injuries are caused when a force acts on the body. If this force is large enough to take a part of the body past its elastic limit, then damage will be done to that part of the body. If the body inside the car is not restrained by seatbelts, then when the car comes to a sudden halt, there isn't anything to stop the body, so it will continue at the speed that it was travelling. The collision of the body with parts of the interior of the car at this speed are most likely to cause a lot of injury.
The impulse equation F = mv, indicates that for a fixed change in momentum (ie. the car and body come to rest) then the larger the smaller F will be.

Question 137
5M - 5m
The total momentum is given by the sum of the individual momentums. You need to consider the vector nature of momentum.
Assume that to the left is positive.
 = M × 5 - m x 5
 = 5M - 5m
The other way of considering this is to use the final velocity. As momentum is conserved, the final momentum is also equal to the initial momentum. Pf = (M + m) × 1.5

Question 138
56Ns
Use F= m.  F = 7.0  8.0 = 56Ns

Question 139
The airbag is designed to increase the time of the collision. It expands rapidly and is already deflating by the time the head comes into contact with it. This deflating bag increases the