A block with mass m1 = 0.600 kg is released from rest on a frictionless track at a distance h1 = 2.70 m above the top of a table. It then collides elastically with an object having mass m2 = 1.20 kg that is initially at rest on the table, as shown in the figure below. (a) Determine the velocities of the two objects just after the collision. (Assume the positive direction is to the right. Indicate the direction with the signs of your answers.(a) Determine the velocities of the two objects just after the collision. (Assume the positive direction is to the right. Indicate the direction with the signs of your answers.(a) Determine the velocities of the two objects just after the collision. (Assume the positive direction is to the right. Indicate the direction with the signs of your answers.



Answer :

The potential energy of the block is 15.6042 Joules. This energy converts to kinetic energy after the collision.

Applying conservation of momentum: 0.600 * v1 + 1.20 * v2 = 0.

Let's solve for v1 and v2 using these equations. Starting with the conservation of energy:

15.6042 = 0.5 * 0.600 * v1^2 + 0.5 * 1.20 * v2^2

Simplifying, we get:

15.6042 = 0.300 * v1^2 + 0.600 * v2^2

Next, we use the conservation of momentum equation:

0.600 * v1 + 1.20 * v2 = 0

Let's solve the system of equations.

From the conservation of momentum equation:

0.600 * v1 + 1.20 * v2 = 0

We can express v2 in terms of v1:

v2 = -0.500 * v1

Now, substitute v2 in terms of v1 into the conservation of energy equation:

15.6042 = 0.300 * v1^2 + 0.600 * (-0.500 * v1)^2

Simplify the equation:

15.6042 = 0.300 * v1^2 + 0.600 * 0.250 * v1^2

15.6042 = 0.300 * v1^2 + 0.150 * v1^2

15.6042 = 0.450 * v1^2

Now, solve for v1:

v1^2 = 15.6042 / 0.450

v1^2 = 34.676

v1 = ±√34.676

v1 ≈ ±5.887 m/s

Since v1 is positive, we take the positive value:

v1 ≈ 5.887 m/s

Now, use v2 = -0.500 * v1 to find v2:

v2 = -0.500 * 5.887

v2 ≈ -2.944 m/s

So, after the collision, the block's velocity is approximately 5.887 m/s to the right, and the object's velocity on the table is approximately 2.944 m/s to the left.

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