-A block of mass m = 2.2 kg,moving on frictionless surface with a speed v_i = 9.1 m/s,makes a perfectly elastic collision with a block of mass M at rest.After the collision,the 2.2 kg block recoils with a speed of v_f = 0.7 m/s.In the figure,the mass M is closest to:

Question

-A block of mass m = 2.2 kg,moving on frictionless surface with a speed v<sub>i</sub> = 9.1 m/s,makes a perfectly elastic collision with a block of mass M at rest.After the collision,the 2.2 kg block recoils with a speed of v<sub>f</sub> = 0.7 m/s.In the figure,the mass M is closest to: A) 2.6 kg B) 18.48 kg C) 2.4 kg D) 31 kg E) 2.2 kg

Quizplus · Accepted Answer

Using the conservation of momentum and kinetic energy for a perfectly elastic collision, we can solve for M. The equation for momentum conservation is: m * vi = m * vf + M * V, where V is the velocity of mass M after the collision. The equation for kinetic energy conservation is: 0.5 * m * vi^2 = 0.5 * m * vf^2 + 0.5 * M * V^2. Solving these equations gives M ≈ 2.6 kg.

-A Block of Mass M = 2