A- if the collision is perfectly elastic, what is the balls speed immediately af

Question

A- if the collision is perfectly elastic, what is the balls speed immediately after the collision? B- what is the maximum compression of the spring? C- Repeat part a for the case of a perfectly inelastic collision.
I’m not sure what I’m doing wrong but I keep getting the wrong answer. A 150 g block on a frictionless table is firmly attached to on . The other e end of a spring with k = 29 N/m end of the spring is anchored to the wall. A 28 g ball is thrown horizontally toward the block with a speed of 5.0 m/s

Explanation / Answer

Let the stationary block be mB and the ball be mA Then

vA (after) = (mA - mB)/(mA + mB)*vA (before) = (0.028-0.150)/(0.028+0.150)*5.0m/s = 3.426m/s

B)

vB = 2*mA/(mA+mB)*vA(before) = 2*0.028/(0.028+0.150)*5.0 = 1.573m/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(0.150kg*(1.573m/s)^2/29N/m) = 0.1131m

A For a completely inelastic collision V after = mA*vA/(mA+mB) = 0.028kg*5.0m/s/(0.028+0.150)kg = 0.7865 m/s

Now 1/2*(mA+mB)*V^2 = 1/2*k*x^2 so x = sqrt((mA+mB)*V^2/k) = sqrt((0.028+0.150)*0.7865^2/29) = 0.0616 m

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