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.
