Block A has a mass of 0.5kg, and block B has a mass of 2kg. Block is is released at a height of 0.75 meters above B. The coefficient of restitution between A and B is 0.7. If, determine the distance the spring deforms before the block comes to rest again.

0.075 m

Explanation:

The picture of the problem is missing: find it in attachment.

At first, block A is released at a distance of

h = 0.75 m

above block B. According to the law of conservation of energy, its initial potential energy is converted into kinetic energy, so we can write:

where

is the acceleration due to gravity

is the mass of the block

is the speed of the block A just before touching block B

Solving for the speed,

Then, block A collides with block B. The coefficient of restitution in the collision is given by:

where:

e = 0.7 is the coefficient of restitution in this case

is the final velocity of block B

is the final velocity of block A

is the initial velocity of block B

Solving,

Re-arranging it,

(1)

Also, the total momentum must be conserved, so we can write:

where

And substituting (1) and all the other values,

This is the velocity of block B after the collision. Then, its kinetic energy is converted into elastic potential energy of the spring when it comes to rest, according to

where

k = 600 N/m is the spring constant

x is the compression of the spring

And solving for x,

lol

explanation:

answer and explanation;