Answer to Question #161281 in Physics for An

Question

A 3-kg block moving to the right on a frictionless table at 4 m/s makes a head on collision with another 3-kg block moving 5 m/s to the left.

*If the collision is completely inelastic, find the final velocity of the blocks

6.

*If the collision is completely elastic, find the final velocities of the blocks.Required to answer. Single choice.

*If half the initial energy is dissipated in the collision, find the final velocities of the blocks.Required to answer. Single choice.

Answer 1

A. If the collision is completely inelastic, find the final velocity of the blocks using momentum conservation:

$$p_i=p_f,\\ mv-Mu=(m+M)s, \\\space\\ s=\frac{mv-Mu}{m+M}=\frac{3\cdot4-3\cdot5}{3+3}=-0.5\text{ m/s}.$$

The two blocks will move to the left at 0.5 m/s as one.

B. If the collision is completely elastic, find the final velocities of the blocks.

In this scenario, we will also introduce conservation of energy:

$$p_i=p_f,\\ mv_1-Mu_1=mv_2+Mu_2, \\\space\\ \frac12mv_1^2+\frac12Mu_1^2=\frac12mv_2^2+\frac12Mu_2^2.$$

In this system of equations, we have two undefined values, namely, $v_2, u_2.$ The solution is

$$v_2=-5\text{ m/s},\\ u_2=4\text{ m/s}.$$

C. If half the initial energy is dissipated in the collision, find the final velocities of the blocks. Use the reasonings from part B, but include the lost energy condition: just 50% of the energy goes for final kinetic energy of motion.

$$p_i=p_f,\\ mv_1-Mu_1=mv_2+Mu_2, \\\space\\ \frac12\bigg(\frac12mv_1^2+\frac12Mu_1^2\bigg)=\frac12mv_2^2+\frac12Mu_2^2.$$

The solution:

$$v_2=-3.66\text{ m/s}, \\ u_2=2.66\text{ m/s}.$$