Conservation of Momentum Problems

Using the law of conservation of momentum, we can set up the equation: (2.0 kg)(4.0 m/s) + (3.0 kg)(-2.0 m/s) = (5.0 kg)v, where v is the final velocity. Solving for v, we get v = 0.8 m/s.

Since the gymnast is on a frictionless floor, there is no external force acting on the system. The total momentum of the gymnast and floor is conserved. Since the initial momentum of the floor is 0, the final momentum of the floor must be equal to the initial momentum of the gymnast. Therefore, the velocity of the floor is -0.04 m/s.

Using the law of conservation of momentum, we can set up the equation: (0.5 kg)(5.0 m/s) = (0.5 kg + 0.2 kg)v, where v is the final velocity. Solving for v, we get v = 3.33 m/s.

<p>Since the objects bounce off each other, we can use the law of conservation of momentum to find the velocity of the 10 kg object after the collision. Let's assume the velocity of the 10 kg object after the collision is v. Then, the velocity of the 5 kg object after the collision is -(3.0 m/s - v). Using the law of conservation of momentum, we can set up the equation: (10 kg)(3.0 m/s) = (10 kg)v + (5 kg)(-(3.0 m/s - v)). Solving for v, we get v = 2.0 m/s.</p> Signup and view all the answers

<p>The initial momentum of the system is (20 kg)(2.0 m/s) = 40 kg m/s. The final momentum of the system is (20 kg + 10 kg)(v), where v is the final velocity. Using the law of conservation of momentum, we can set up the equation: (20 kg)(2.0 m/s) = (30 kg)v, where v = 1.33 m/s. The final momentum of the system is (30 kg)(1.33 m/s) = 40 kg m/s. Therefore, the percentage change in momentum is 0%.</p> Signup and view all the answers