Answer to Question #159832 in Physics for Garison

Question

A block of mass 4 kg is released from rest at the top of the curved frictionless

ramp shown below. The block slides down the ramp and collides with a 6 kg block at

rest at the bottom of the incline. The 6 kg block moves to the right at a speed

12.8 m/s immediately after the collision.  The blocks do not stick together.

(a) The larger block slides a distance 15 m before coming to rest. Determine the

value of the coefficient of kinetic friction between the 6 kg block and the 

surface on which it slides. 

(b) Indicate whether the collision between the two blocks is elastic or inelastic. 

Justify your answer.

Answer 1

a) According to Work-Kinetic Energy Theorem, we have:

$$\Delta KE=W,$$ $$KE_f-KE_i=-F_{fr}d,$$ $$0-\dfrac{1}{2}mv_i^2=-\mu mgd,$$ $$\mu=-\dfrac{-v_i^2}{2gd}=-\dfrac{-(12.8\ \dfrac{m}{s})^2}{2\cdot9.8\ \dfrac{m}{s^2}\cdot 15\ m}=0.56$$

b) Let's first find the final velocity of the smaller block at the moment of collision with the larger block being at rest. We can find it from the law of conservation of energy. Unfortunately, there is no picture attached and we don't know the height of the curfed frictionless ramp. Let's assume that the height of the ramp equals 3.5 meters. Than, we get:

$$PE=KE,$$ $$mgh=\dfrac{1}{2}mv^2,$$ $$v=\sqrt{2gh}=\sqrt{2\cdot9.8\ \dfrac{m}{s^2}\cdot3.5\ m}=8.28\ \dfrac{m}{s}.$$

Let's find the velocity of the smaller block after the collision from the law of conservation of energy:

$$m_1v_{1i}=m_1v_{1f}+m_2v_{2f},$$ $$v_{1f}=\dfrac{m_1v_{1i}-m_2v_{2f}}{m_1},$$ $$v_{1f}=\dfrac{4\ kg\cdot8.28\ \dfrac{m}{s}-6\ kg\cdot12.8\ \dfrac{m}{s}}{4\ kg}=-10.92\ \dfrac{m}{s}.$$

Let's check that the linear momentum is conserved:

$$p_i=m_1v_{1i}=4\ kg\cdot8.28\ \dfrac{m}{s}=33.12\ \dfrac{kgm}{s},$$

$p_f=m_1v_{1f}+m_2v_{2f}=4\ kg\cdot(-10.92\ \dfrac{m}{s})+6\ kg\cdot12.8\ \dfrac{m}{s}=33.12\ \dfrac{kgm}{s}.$

Thus, $p_i=p_f$ and linear momentum is conserved.

Now. let's check whether the kinetic energy conserved or not.

$$K_i=\dfrac{1}{2}m_1v_{1i}^2=\dfrac{1}{2}\cdot4\ kg\cdot(8.28\ \dfrac{m}{s})^2=137.11\ J,$$ $$K_f=\dfrac{1}{2}m_1v_{1f}^2+\dfrac{1}{2}m_2v_{2f}^2,$$ $$K_f=\dfrac{1}{2}\cdot4\ kg\cdot(-10.92\ \dfrac{m}{s})^2+\dfrac{1}{2}\cdot6\ kg\cdot(12.8\ \dfrac{m}{s})^2=730\ J.$$

Thus, $K_i \not= K_f$ and kinetic energy isn't conserved.

Therefore the collision is inelastic.