Question

a block of mass m = 2.90 kg starts from rest and slides down a 30.0° incline which is 3.60 m high. at the bottom, it strikes a block of mass m = 8.00 kg which is at rest on a horizontal surface (figure 1). (assume a smooth transition at the bottom of the incline.) the collision is elastic, and friction can be ignored.

part a
determine the speed of the block with mass m = 2.90 kg after the collision.
express your answer to three significant figures and include the appropriate units.

part b
determine the speed of the block with mass m = 8.00 kg after the collision.
express your answer to three significant figures and include the appropriate units.

Explanation:

Step1: Find speed of block m before collision

Use conservation of energy: gravitational potential energy converts to kinetic energy.
$$mgh = \frac{1}{2}mv_0^2$$
$$v_0 = \sqrt{2gh} = \sqrt{2\times9.81\times3.60} \approx 8.406 \, \text{m/s}$$

Step2: Recall elastic collision formulas

For elastic collision, final velocities are:
$$v_m = \frac{m-M}{m+M}v_0$$
$$v_M = \frac{2m}{m+M}v_0$$

Step3: Calculate $v_m$ (Part A)

Substitute $m=2.90\ \text{kg}$, $M=8.00\ \text{kg}$, $v_0\approx8.406\ \text{m/s}$
$$v_m = \frac{2.90-8.00}{2.90+8.00}\times8.406 = \frac{-5.10}{10.90}\times8.406 \approx -3.91 \, \text{m/s}$$
The negative sign indicates direction opposite to initial motion.

Step4: Calculate $v_M$ (Part B)

$$v_M = \frac{2\times2.90}{2.90+8.00}\times8.406 = \frac{5.80}{10.90}\times8.406 \approx 4.49 \, \text{m/s}$$

Answer:

Part A: $v_m = -3.91$ m/s
Part B: $v_M = 4.49$ m/s