Two boys standing on carts that move along a horizontal frictionless surface. The mass of each boy-cart combination is 100 kg, and each moves initially with the velocities of: Boy A 7cm/sec and Boy B: 10cm/sec. The boy on cart A throws a 2kg ball to the boy on Cart B (assume it travels horizontally). The boy on cart B then throws the ball at 15m/sec in the forward direction and finds that his cart has come completly to rest. At what speed must the boy on cart A have thrown the ball in order to make this happen?
To solve this problem, we need to apply the principle of conservation of linear momentum. According to this principle, the total linear momentum before and after the interaction remains constant if no external forces are acting on the system.
Let's denote the velocity of the boy on cart A after throwing the ball as vA and the final velocity of the boy on cart B after receiving the ball as vB.
Before the interaction, the total initial momentum of the system is the sum of the momenta of boy A and boy B, given by:
Initial momentum = (mass of Boy A) * (velocity of Boy A) + (mass of Boy B) * (velocity of Boy B)
Initial momentum = (100 kg) * (0.07 m/s) + (100 kg) * (0.10 m/s)
After the interaction, boy B throws the ball with a velocity of 15 m/s. As a result, the ball imparts an equal and opposite momentum to boy B, causing him to come to rest. Therefore, the momentum of boy B after the interaction is zero.
Using the conservation of linear momentum, we can set up the equation:
Initial momentum = Final momentum
(100 kg) * (0.07 m/s) + (100 kg) * (0.10 m/s) = (100 kg) * (vA) + (100 kg) * (0 m/s)
Simplifying this equation, we get:
(7 + 10) kg·m/s = 100 kg·m/s * vA
17 kg·m/s = 100 kg·m/s * vA
To find vA, divide both sides of the equation by 100 kg·m/s:
vA = 17 kg·m/s / 100 kg·m/s
vA = 0.17 m/s
Therefore, the boy on cart A must throw the ball with a velocity of 0.17 m/s to bring the boy on cart B to a complete stop.