Answer to Question #97954 in Mechanics | Relativity for Abhi

Question

A 60.0 kg person running at an initial speed of 4.00 m/s jumps onto a 120 kg cart
initially at rest as shown below. The person slides on the cart’s top surface and finally comes to rest relative to the cart. The coefficient of kinetic friction between the person and the cart is μk =
0.400. Friction between the cart and the ground can be ignored.
(g) Find the displacement of the cart relative to the ground while the person is sliding on the cart.
(h) Find the change in kinetic energy of the person.
(i) Find the change in kinetic energy of the cart.

Answer 1

From the conservation of momentum:

$$mv=(m+M)V\to V=v\frac{m}{m+M}$$

g)

$$\mu mgd =K'-K=0.5mv^2-0.5(m+M)V^2$$

$$\mu mgd =0.5mv^2-0.5(m+M)\left(v\frac{m}{m+M}\right)^2$$

$$2\mu gd =v^2\left(1-\left(\frac{m}{m+M}\right)\right)$$

$$2(0.4)(9.8)d =4^2\left(1-\left(\frac{60}{60+120}\right)\right)$$

$$d=1.36\ m$$

h)

$$\Delta K_p=0.5mV^2-0.5mv^2=0.5m\left(\left(v\frac{m}{m+M}\right)^2-v^2\right)$$

$$\Delta K_p=0.5mv^2\left(\left(\frac{m}{m+M}\right)^2-1\right)$$ $$\Delta K_p=0.5(60)(4)^2\left(\left(\frac{60}{60+120}\right)^2-1\right)=-427\ J$$

i)

$$\Delta K_c=0.5mV^2=0.5m\left(\left(v\frac{m}{m+M}\right)^2\right)$$

$$\Delta K_c=0.5(60)(4)^2\left(\left(\frac{60}{60+120}\right)^2\right)=53.3\ J$$