Question

A spring of negligible mass has a force constant k=1600N/m. You place the spring vertically with one end on the floor. You then drop a 1.20-kg book onto it from a height of 0.800 m above the top of the spring. The magnitude for the impulse needed to stop the falling book by the compressed spring would?

o Increase by the same amount

o Increase by 4 times as much if the height is doubled

o Double if the height is increased by 4 times as much

o None of the above

Answer 1

We know that the impulse is equal to change in momentum
= Initial momentum - final momentum
= mV- 0
where m is mass of book and V is velocity just before reaching the spring.
Say intial energy of book = mgh
Final energy of book = (1/2)mV²
applying the energy conservation
(1/2)mV² = mgh
V = (2gh)^1/2
Now the magnitude of impulse = m*(2gh)^1/2
If we increased the height 4 times i.e. H = 4h
Impulse = m*(2gH)^1/2 = m*(2g*(4h))^1/2 = 2*(m)*(2gh)^1/2
Hence the magnitude of the impulse becomes double.
so the correct option would be C.
i.e. Double if the height is increased by 4 times.