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?
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To solve this problem, we can use the law of conservation of momentum. According to this law, the total momentum before the boy jumps off should be equal to the total momentum after the boy jumps off.
The momentum of an object is calculated by multiplying its mass by its velocity.
Given information:
- Mass of the cart (m1) = 50 kg
- Initial velocity of the cart (v1) = 2.0 m/s
- Mass of the child (m2) = 70 kg
- Final velocity of the child (v2) = 0 m/s
Let's assume the velocity of the cart after the boy jumped off is v3.
Now, let's calculate the momentum before and after the boy jumps off:
Before the boy jumps:
Momentum of the cart (m1 * v1) + Momentum of the child (m2 * v1) = Total initial momentum
After the boy jumps:
Momentum of the cart (m1 * v3) + Momentum of the child (m2 * v2) = Total final momentum
Using the conservation of momentum, we can equate the two equations:
m1 * v1 + m2 * v1 = m1 * v3 + m2 * v2
Substituting the given values:
50 kg * 2.0 m/s + 70 kg * 2.0 m/s = 50 kg * v3 + 70 kg * 0 m/s
Next, simplify the equation:
100 kg⋅m/s + 140 kg⋅m/s = 50 kg * v3
Now, solve for v3:
240 kg⋅m/s = 50 kg * v3
Divide both sides of the equation by 50 kg:
4.8 m/s = v3
Therefore, the velocity of the cart after the boy jumps off is 4.8 m/s.