Question

Please label work... if you may ..thank you!

111 ·.. A 1.00-kg block (mass m) and a second block (mass M) are both initially at rest on a frictionless inclined plane (Figure 8-56). Mass M rests against a spring that has a force con- stant of 11.0 kN/m. The distance along the plane between the two blocks is 4.00 m. The 1.00-kg block is released, making an elastic collision with the larger block. The 1.00-kg block then re- bounds a distance of 2.56 m back up the inclined plane. The block of mass M momentarily comes to rest 4.00 cm from its ini tial position. Find M. SSM 1m 4.00 m 30°

Answer 1

applying momentum conservation during collision,

(1 x 6.26) + (0) = (1 x -5.01) + (M)(1.25)

M = 9.02 kg ...........Ans

Using energy equation,

W_spring + W_gravity = KEf - KEi

-k d^2 /2 + M g d sin30 = 0 - M v^2 /2

- (11000)(0.04^2)/2 + (M x 9.8 x 0.04 x sin30) = - M(1.25^2)/2

-8.8 + 0.196M =- 0.78125M

M = 9 kg