momentum conservation
initial=final
100*3+50*(-8)=100*0 + 50*V
I agree with your answer.
I think the answer is -2 m/s.
initial=final
100*3+50*(-8)=100*0 + 50*V
I agree with your answer.
The momentum of an object is defined as the product of its mass and velocity. The principle of conservation of momentum states that the total momentum of a closed system remains constant before and after a collision.
In this case, the closed system consists of Object 1 and Object 2.
Before the collision, the total momentum of the system is given by:
Total momentum before = (mass of Object 1 * velocity of Object 1) + (mass of Object 2 * velocity of Object 2)
Let's calculate the initial momentum of the system:
Total momentum before = (100 kg * 3 m/s) + (50 kg * (-8 m/s))
Total momentum before = 300 kg路m/s - 400 kg路m/s
Total momentum before = -100 kg路m/s
Since the system is closed and there are no external forces acting on it, the total momentum after the collision will also be -100 kg路m/s.
After the collision, Object 1 stops moving, so its final velocity is 0 m/s.
Let's denote the final velocity of Object 2 as v2 (to be determined).
The total momentum after the collision is given by:
Total momentum after = (mass of Object 1 * final velocity of Object 1) + (mass of Object 2 * final velocity of Object 2)
Since the final velocity of Object 1 is 0 m/s, we can simplify the equation to:
Total momentum after = (mass of Object 2 * final velocity of Object 2)
Equating the total momentum after the collision to the initial momentum, we have:
-100 kg路m/s = 0 kg * 0 m/s + 50 kg * final velocity of Object 2
Solving for the final velocity of Object 2:
final velocity of Object 2 = (-100 kg路m/s) / (50 kg)
final velocity of Object 2 = -2 m/s
Therefore, the second object would be moving at -2 m/s after the collision. Your answer of -2 m/s is correct!