Question

A 1.5−kg block moves to the right and hit a 2-kg block that is initially at rest on a horizontal surface. After the collision, the 1.5−kg block slides to the left with a speed of −5 m/s, and the 2−kg block slides to the right for 7.5 m before coming to rest. The coefficient of kinetic friction between the block and the surface is 0.6. (a) What is the frictional force exerted on the 2 kg-block when it is sliding? (b) What is the speed of the 2−kg block immediately after its being hit by the 1.5−kg block? (c) What was the speed of the 1.5−kg immediately before the collision?

Answer 1

Part (a)

`F_f=\mu_k F_N=\mu_k mg=(0.6)(9.81)(2)=11.772\text{ N}`

Part (b)

`F=ma \implies a=(F)/(m) \\ \\ \\ v_f^2=v_0^2+2a \Delta x \\ v_0^2=-2(F/m)\Delta x \\ \\ v_0=\sqrt{(-2F \Delta x)/(m)} \\ \\ v_0=\sqrt{(-2(-11.772)(7.5))/(2)}=9.40 \text{ m/s}`

Part (c)

`m_1 v_(1,0)+m_2 v_(2,0)=m_1 v_(1,f)+m_2 v_(2,f) \\ \\ 1.5v_(1,0)+2(0)=1.5(-5)+2(9.40) \\ \\ v_(1,0)=7.53 \text{ m/s}`