A 50. kg cart is moving across a frictionless floor at 2.0 m/s. A 70. kg child, riding in the cart, jumps off so that the child hits the floor with zero velocity. What was the velocity of the cart after the boy jumped?
(M1+M2)V1=M1Vf+M2*0
conservation of momentum.
4.8m/s
To solve this problem, we can use the principle of conservation of linear momentum. The total momentum before the boy jumps off should be equal to the total momentum after the boy jumps off.
Let's denote the velocity of the cart after the boy jumps off as Vc (final velocity of the cart) and the velocity of the child just before he jumps off as Vb (initial velocity of the child).
Given:
Mass of the cart (m1) = 50 kg
Velocity of the cart (V1) = 2.0 m/s
Mass of the child (m2) = 70 kg
Velocity of the child (V2) = ?
Using the conservation of linear momentum, we have:
(mass of the cart * velocity of the cart) + (mass of the child * velocity of the child) = (mass of the cart * final velocity of the cart) + (mass of the child * 0)
(50 kg * 2.0 m/s) + (70 kg * V2) = (50 kg * Vc) + (70 kg * 0)
100 kg m/s + 70 kg * V2 = 50 kg * Vc
Now, substitute the given values into the equation:
100 kg m/s + 70 kg * V2 = 50 kg * Vc
100 kg m/s + 70 kg * V2 = 50 kg * Vc
To find the final velocity of the cart (Vc), we need to know the velocity of the child just before he jumps off (V2). Unfortunately, this information is not given in the question. Without the velocity of the child just before jumping off, we cannot determine the final velocity of the cart.
To determine the velocity of the cart after the boy jumps off, we can use the principle of conservation of momentum. According to this principle, the total momentum before the boy jumps off should equal the total momentum after the boy jumps off.
The momentum of an object is calculated by multiplying its mass by its velocity. Let's define the positive direction as the initial direction of motion of the cart.
Initially, the cart has a mass of 50 kg and a velocity of 2.0 m/s, so its initial momentum is:
Initial momentum of the cart = mass of the cart * velocity of the cart
= 50 kg * 2.0 m/s
= 100 kg⋅m/s (in the positive direction)
When the boy jumps off, the cart experiences no external forces, so the total momentum after the boy jumps off is zero. This is because the boy's momentum in one direction is canceled out by the cart's momentum in the opposite direction.
Let's assume the final velocity of the cart is v (in the positive direction). The boy's mass is 70 kg, and since he hits the floor with zero velocity, his momentum after jumping off is zero.
According to the principle of conservation of momentum, the total momentum before the boy jumps off (100 kg⋅m/s) should equal the total momentum after the boy jumps off (zero).
Thus, we can set up the equation:
Initial momentum of the cart = Final momentum of the cart
100 kg⋅m/s = (50 kg + 70 kg) * v
Simplifying the equation:
100 kg⋅m/s = 120 kg * v
Dividing both sides of the equation by 120 kg:
v = (100 kg⋅m/s) / 120 kg
v ≈ 0.833 m/s
Hence, the velocity of the cart after the boy jumps off is approximately 0.833 m/s in the positive direction.