Question

A 465 g block slides along a frictionless surface at a speed of 0.35 m/s. It runs into a horizontal massless spring with spring constant 54 N/m that extends outward from a wall. It compresses the spring, then is pushed back in the opposite direction by the spring, eventually losing contact with the spring.
a) How long does the block remain in contact with the spring?
b) How would your answer change if the block's initial speed was doubled?

Answer 1

Answer:

a) The duration, during which the block remain in contact with the spring is 0.29 s

b) The period of the simple harmonic oscillatory motion depends only on the mass and spring constant, therefore when the speed is doubled, the duration of contact remains the same as 0.29 s.

Explanation:

Mass of the block = 465 g

Surface speed = 0.35 m/s

Spring constant , k = 54 N/m

$T = 2\times \pi \times \sqrt{\frac{m}{K} } = 2\times \pi \times \sqrt{\frac{0.465}{54} }$  = 0.58 s

a) Since the period for the oscillatory motion is 0.58 s, then the time when the block and spring remain in contact is T/2 = 0.29 s

b) When the speed is doubled, we have

$T = 2\times \pi \times \sqrt{\frac{m}{K} }$

Therefore, since T is only dependent on the mass, m and the  spring constant, K, then the time it takes when the speed is doubled remain as

T /2 = 0.29 s