Collision between two blocks on a table

[Clutch306]

In the attached figure, block A of mass 2.0kg slides on a horizontal frictionless table with velocity of 3.0m/s and collides with block B of mass 3.0kg initially at rest on the edge of the table. Block B moves with a horizontal velocity of 2.4m/s, just after the collision.

(a) Use conversation of momentum to find the velocity of Block A, after the collision
(b) Is the collision elastic? (Hint: Verify K.E. of the block before & after collision)
(c) What is the kinetic energy of block B, just before it strikes the floor 0.40m below the table's surface? (Hint use conservation of M.E.)

 

[recon]

At least show us the work you've done. If this is taken from a textbook, have a look at the textbook examples. I'm sure there are plenty.

 

[M98Ranger]

(a) According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Therefore, we can set up the following equation:

m1v1 + m2v2 = m1v1' + m2v2'

Where m1 and v1 are the mass and velocity of block A before the collision, m2 and v2 are the mass and velocity of block B before the collision, and v1' and v2' are the velocities of block A and B after the collision.

Plugging in the given values, we get:

(2.0kg)(3.0m/s) + (3.0kg)(0m/s) = (2.0kg)v1' + (3.0kg)(2.4m/s)

Solving for v1', we get:

v1' = 1.8m/s

Therefore, the velocity of block A after the collision is 1.8m/s.

(b) To determine if the collision is elastic, we need to compare the kinetic energy of the blocks before and after the collision. The kinetic energy before the collision is given by:

KE = 1/2mv^2

For block A, KE = (1/2)(2.0kg)(3.0m/s)^2 = 9J
For block B, KE = (1/2)(3.0kg)(0m/s)^2 = 0J

The total kinetic energy before the collision is 9J.

After the collision, the kinetic energy for block A is given by:

KE = (1/2)(2.0kg)(1.8m/s)^2 = 3.24J

The kinetic energy for block B is given by:

KE = (1/2)(3.0kg)(2.4m/s)^2 = 8.64J

The total kinetic energy after the collision is 12.88J.

Since the total kinetic energy before and after the collision are not equal, we can conclude that the collision is not elastic.

(c) To find the kinetic energy of block B just before it strikes the floor, we can use the conservation of mechanical energy. Before the collision, the total mechanical energy is given by the potential energy of block B, which is equal to its mass multiplied by the acceleration due to gravity (9.8m/s^2) multiplied