Question

A four-wheel cart of mass M = 95 kg is moving along a horizontal surface with a constant velocity V = 3.5 m/s relative to the ground. A person of mass m1 = 65 kg carrying a backpack of m2 = 8 kg runs and catches up to the cart, and then jumps onto the cart. Just before landing on the cart, the person is moving parallel to the ground and the velocity of the center of mass of the system including the person, backpack and cart is VCM = 5 m/s.

What is the speed of the person just before landing on the cart?

v₀ = 5.3 m/s

v₀ = 12 m/s

v₀ = 0.45 m/s

v₀ = 7 m/s

v₀ = 8.8 m/s

2)

What is the horizontal momentum of the person after landing on the cart?

p_f = 325 kg m/s

p_f = 455 kg m/s

p_f = 228 kg m/s

3)

Compare the total kinetic energy of the system including the person, backpack and cart before the person has landed on the cart to after.

KE_before = KE_after

KE_before > KE_after

KE_before < KE_after

4)

The person now holds the backpack off the back of the cart and lets go. The backpack falls to the ground. What happens to the speed of the cart when the backpack is dropped?

increases

decreases

stays the same

(Note: Answers are D, A, B, C. Please show work and reasoning.)

Answer 1

A) Velocity of center of mass of number of masses is
V_cm = sum of momentum of masses / sum of masses   ...1
Momentum of mass = mass* velocity

Hence
5 = ( 95*3.5 + (65+8) V ) / (95+65+8)
V is velocity of person with back pack
V = 7 m/s

B) Velocity of center of mass remains same during collision.
    If number of masses move with same velocity, that s also velcoity of center of mass ( from first equation)

As person lands on cart, all three, cart , person and back pack move with same velocity. That is velocity of center of mass. This is same as velocity of center of mass before collision. Hence velocity of person is now 5 m/s.
Momentum of person = 65*5 = 325 kg m/sec

C) Kinetic energy = mv²/2
KE(before) = 95*3.5²/2 + (65+8)*7² /2
KE(after) = (95+65+8)*5² /2
Hence KE(before) > KE(after)

D) Again there is no change in velocity of center of mass in horizontal direction. As back pack is dropped, there is no change in it's horizontal velocity, that is it remains 5m/s, there is no change in velocity of cart + person
Hence speed of cart stays same.