physics question

Question

A block of known mass M=1.5kg is at rest at the base of a ramp. A second identical block of mass M=1.5kg travels at a known, constant velocity v0 = 15.7m/s,. The block that travels at a constant speed collides with and sticks to the block at the base of the ramp. Both blocks slide up the ramp and travel with an unknown velocity vR at the top of the ramp, which is 2.5m higher that the base. All frictional forces are considered to be negligible. What is the velocity vR of the combined blocks at the top of the ramp? Ans: 3.55 m/s

Answer 1

Since the two masses stick together after the collision, the collision is inelastic (the kinetic energy is not conserved). However, there are no external forces (gravity and the normal force cancel each other out) at the moment of collision, thus conserving momentum.

Mv₀ = 2Mv_x, with v_x the speed of both masses after the collision, before going on the ramp.

After the collision, the only forces working on the system (the two masses together) are conservative (gravity and normal force, no frictional forces). This means no energy is converted into heat, sound or deformation, thus conserving mechanical energy (KE + PE = constant).

2Mv_x² / 2 = 2Mv_R² / 2 + 2Mgh (see that you understand how I got to this equation! Let me know otherwise.)

Solve for v_R and you get your solution.