Question

A 13.0-g wad of sticky clay is hurled horizontally at a 110-g wooden block initially at rest on a horizontal surface. The clay sticks to the block. After impact, the block slides 7.50 m before coming to rest. If the coefficient of friction between block and surface is 0.650, what was the speed of the clay immediately before impact?

Answer 1

Answer:

$v_{ic}=92.53 m/s$

Explanation:

We need to apply conservation of momentum and energy to solve this problem.

Conservation of momentum

$p_{i}=p_{f}$

$m_{c}v_{ic}=(m_{c}+m_{w})V$ (1)

• m(c) is the mass of stick clay
• m(w) is the mass of the wooden block
• v(ic) is the initial velocity of clay
• V is the final velocity of the system clay plus wood.

Conservation of total energy

The change in kinetic energy is equal to the change in internal energy, in our case it would be the energy loss due to the friction force. Let's recall the definition of work, it is the dot product between force and displacement, Therefore:

$\Delta E=W$

$\frac{1}{2}(m_{c}+m_{w})V^{2}=F_{friction}*d$

$\frac{1}{2}(m_{c}+m_{w})V^{2}=\mu (m_{c}+m_{w})gd$

We can find V from this equation:

$V=\sqrt{2\mu gd}=\sqrt{2*0.65*9.81*7.5}=9.78 m/s$

Now, let's put V into the equation (1) and find v(ic)

$v_{ic}=\frac{(m_{c}+m_{w})V}{m_{c}}=\frac{123*9.78}{13}=92.53 m/s$

I hope it helps you!