A four-wheel cart of mass M = 95 kg is moving along a horizontal surface with...

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Physics

A four-wheel cart of mass M = 95 kg is moving along a horizontalsurface with a constant velocity V = 3.5 m/s relative to theground. A person of mass m1 = 65 kg carrying a backpack of m2 = 8kg runs and catches up to the cart, and then jumps onto the cart.Just before landing on the cart, the person is moving parallel tothe ground and the velocity of the center of mass of the systemincluding the person, backpack and cart is VCM = 5 m/s.
What is the speed of the person just before landing on thecart?
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 onthe 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 theperson, backpack and cart before the person has landed on the cartto 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 andlets go. The backpack falls to the ground. What happens to thespeed of the cart when the backpack is dropped?
increases
decreases
stays the same
(Note: Answers are D, A, B, C. Please show work andreasoning.)

Answer & Explanation Solved by verified expert

3.5 Ratings (609 Votes)

A Velocity of center of mass of number of masses is Vcm sum of momentum of masses sum of masses 1 Momentum of mass mass velocity Hence 5 9535 658 V 95658 V is velocity of person with back See Answer

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