Two blocks, of mass m and 2m, are initially at rest on a horizontal frictionless surface. A force F is exerted individually on e
1 answer:
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Answer:
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,
