Question

A 140 g block on a frictionless table is firmly attached to one end of a spring with k = 24 N/m . The other end of the spring is anchored to the wall. A 26 g ball is thrown horizontally toward the block with a speed of 6.0 m/s . If the collision is perfectly elastic, what is the ball's speed immediately after the collision? What is the maximum compression of the spring? Repeat part A for the case of a perfectly inelastic collision. Repeat part B for the case of a perfectly inelastic collision.

Answer 1

Applying the eqns of conservation of energy and momentum we can derive the following eqns for the speeds of the objects after an elastic collision. Let the stationary block be mB and the ball be mA Then vA (after) = (mA - mB)/(mA + mB)*vA (before) = [(0.026-0.140)/(0.026+0.140)]*6m/s = -4.12m/s

B) vB = 2*mA/(mA+mB)*vA(before) = 2*0.026/(0.026+0.140)*6 = 1.879m/s

Now this Kinetic energy will be converted to potential energy in the spring So 1/2*mB*vB^2 = 1/2*k*x^2 therefore x = sqrt(mB*vB^2/k) = sqrt(0.140kg*(1.879m/s)^2/24N/m) = 0.143m

A For a completely inelastic collision V after = mA*vA/(mA+mB) = 0.026kg*6m/s/(0.026+0.140)kg = 0.9397m/s

Now 1/2*(mA+mB)*V^2 = 1/2*k*x^2 so x = sqrt((mA+mB)*V^2/k) = sqrt((0.026+0.140)*0.9397^2/24) = 0.0781m