A block is attached to a spring

block-spring-stone

A 15 kg block is firmly attached to a very light horizontal spring (k = 550 N/m) and resting on a smooth horizontal table. Suddenly it is struck horizontally by a 3 kg stone travelling with a velocity of 8 m/s to the right, whereupon the stone rebounds at 2 m/s to the left. Find the maximum distance that the block will compress the spring after the collision.

Solution.

Before the collision only the stone has momentum. After the collision both the stone and the block have momentum, but the total momentum should not change. Let’s use the law of momentum conservation and calculate the velocity of the block just after the collision.

m1v1i + m2v2i = m1v1f + m2v2f

Where m1 is the mass of the block, m2 is the mass of the stone, v1i and v1f are velocities of the block before and after collision respectively, v2i and v2f are velocities of the stone before and after collision respectively.

15 * 0 + 3 * 8 = 15 * v1f – 3 * 2

v1f = 2 m/s

Next we apply the law of conservation of energy

Ek = Usp

where Ek – the kinetic energy of the block and Usp – the potential energy of the spring.

m1 * v1f2 / 2 = k * x2 / 2

Hence,

15 * 22 = 550 * x2

x ≈ 0.33 m

 

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s