A 60.0 kg person running at an initial speed of 4.00 m/s jumps ontoa 120 kg cart
ID: 1662249 • Letter: A
Question
A 60.0 kg person running at an initial speed of 4.00 m/s jumps ontoa 120 kg cart initially at rest. The person slides on thecart's surface and finally comes to rest relative to the cart. The coefficient of kinetic friction between the person andthe cart is 0.400 (friction between the cart and the ground can beneglected).Find (a) the final velocity of the person and cart relative to theground. (b) the friction force acting on the person.
Thanks in advance; will rate soon.
Find (a) the final velocity of the person and cart relative to theground. (b) the friction force acting on the person.
Thanks in advance; will rate soon.
Explanation / Answer
Mass of the person, M1 = 60kg Initial velocity of theperson, U1 = 40 m/s Mass of the cart, M2 =120 kg Initial velocity of thecart, U1 = 0 m/s Coefficient of frictionbetween person and cart's surface, = 0.4 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ (a) From the lawof conservation of momentum, Total finalmomentum of person + cart = Total initial momentum of person andcart ( M1 +M2 ) V = M1 V1 + M2 V2 180 V= 60 * 4 + 0 Final velocity of person + cart, V = 240 / 180 = 1.33 m/s ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ (b) Frictional force acting on the person, f = M1 g = 0.4 * 60 * 9.8 = 235.2 N
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