Exercise on collisions and oscillations

[jones123]

Hi,

I tried to solve this problem myself and I'd like someone to check it :) Thanks already!

A ball is rolling towards a block, which is connected to a spring. Assume no friction occurs. The initial velocity of the ball is 10 m/s. The spring constant is k = 5 N/m. Mass of the ball = 5 kg and mass of the block connected to the spring = 2 kg. After the ball hit the block, the spring starts oscillating harmonically. Calculate:

(a) the final velocity of the ball if it rolls back immediately.
=> i don't really know if you should concern the 'rolling motion' but I used the formula of an elastic collision where m1 = block and m2 = mass
=> v2f = ((m2-m1)v2i + 2m1v1i) / (m1+m2) = (5-2)(10) / 7 = 4,28 m/s

(b) what is the amplitude of the oscillation?
=> conservation of energy : (1/2)mv² = (1/2)kx²
where v = v1f = ((m1-m2)v1i + 2m2v2i) / (m1+m2) = 2(5)(10) / 7 = 14.28 m/s
=> putting this into the formula:
x² = 2(14.28) / 5 => x = 2.38m

(c) the max acceleration of the block

a(max) = kx/m = (5)(2.38)/(2) = 5.95 m/s² or a(max) = Aw² = (2.38)(5/2) = 5.95 m/s²

 

[mfb]

the final velocity of the ball if it rolls back immediately

That is strange. I agree with your approach for an elastic collision, but then the ball does not roll back - it continues to roll in the same direction as before. This also implies that the oscillation of spring+block will hit it again, if the ball does not vanish magically.

For (b), I think you forgot a square for the velocity. This leads to a wrong answer for c, too.

 

[haruspex]

That is strange. I agree with your approach for an elastic collision, but then the ball does not roll back - it continues to roll in the same direction as before. This also implies that the oscillation of spring+block will hit it again, if the ball does not vanish magically.

I agree. Even for an elastic collision, the ball will only roll back if it is less massive than the block. If not completely elastic, it might not roll back even then. And I don't see where it says the collision is elastic anyway, so even if the masses are swapped around there's not enough information.