Consider a collision between two marbles . A 100 g marble moving right at 5.4 m/s strikes a 14 g marble at rest. After the collision, if the 14 g marble moves forward with speed 1.2 m/s, what is the speed, in m/s, of 100 g marble immediately after the collision?
Question
Consider a collision between two marbles . A 100 g marble moving right at 5.4 m/s strikes a 14 g marble at rest. After the collision, if the 14 g marble moves forward with speed 1.2 m/s, what is the speed, in m/s, of 100 g marble immediately after the collision?
Solution
This problem can be solved using the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision.
Step 1: Convert the masses of the marbles to kg (since the standard unit of mass in physics is kg). 100 g = 0.1 kg 14 g = 0.014 kg
Step 2: Calculate the initial momentum before the collision. The formula for momentum is mass times velocity (p=mv). The initial momentum of the 100 g marble is (0.1 kg)(5.4 m/s) = 0.54 kgm/s. The initial momentum of the 14 g marble is (0.014 kg)(0 m/s) = 0 kgm/s. So, the total initial momentum is 0.54 kgm/s + 0 kgm/s = 0.54 kg*m/s.
Step 3: Calculate the final momentum after the collision. We know the 14 g marble moves forward with speed 1.2 m/s, so its final momentum is (0.014 kg)(1.2 m/s) = 0.0168 kg*m/s. We don't know the final velocity of the 100 g marble, so we'll call it v.
Step 4: Set the total initial momentum equal to the total final momentum and solve for v. 0.54 kgm/s = (0.1 kg)v + 0.0168 kgm/s Subtract 0.0168 kgm/s from both sides to isolate v: 0.5232 kgm/s = (0.1 kg)v Divide both sides by 0.1 kg to solve for v: v = 5.232 m/s
So, the speed of the 100 g marble immediately after the collision is approximately 5.232 m/s.
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