A ball is attached to one end of a wire, the other end being fastened to the ceiling. The wire is held horizontal, and the ball is released from rest (see the drawing). It swings downward and strikes a block initially at rest on a horizontal frictionless surface. Air resistance is negligible, and the collision is elastic. The masses of the ball and block are, respectively, 1.7 kg and 2.39 kg, and the length of the wire is 1.22 m. Find the velocity (magnitude and direction) of the ball (a) just before the collision, and (b) just after the collision.
To find the velocity of the ball just before and just after the collision, we can break down the problem into two parts:
Part a: Finding the velocity just before the collision
1. Determine the potential energy of the ball at its highest point: Since the ball is released from rest, all of its initial potential energy is converted to kinetic energy at its highest point. Therefore, we can equate potential energy to kinetic energy:
mgh = (1/2)mv^2, where m is the mass of the ball, g is the acceleration due to gravity, h is the height of the ball (length of the wire), and v is the velocity.
Plugging in the given values:
(1.7 kg)(9.8 m/s^2)(1.22 m) = (1/2)(1.7 kg)v^2
Simplifying the equation:
20.044 Nm = 0.85 kgv^2
Solving for v:
v = √(20.044 Nm / 0.85 kg) ≈ 5.32 m/s
Therefore, the velocity of the ball just before the collision is approximately 5.32 m/s downwards.
Part b: Finding the velocity just after the collision
1. Use the principle of conservation of momentum: In an elastic collision, both momentum and kinetic energy is conserved. Therefore, we can equate the initial momentum to the final momentum.
Initial momentum = ball momentum + block momentum
Initial momentum of the ball (before collision) = ball mass x ball velocity (magnitude and direction)
Block momentum is 0 since it is initially at rest.
Therefore, initial momentum = (1.7 kg) x (-5.32 m/s) = -9.044 kg·m/s
2. Calculate the final momentum: Since the collision is elastic, the total momentum before and after the collision must remain the same.
Final momentum = ball momentum (just after collision) + block momentum (just after collision)
Let v' be the velocity of the ball just after the collision (magnitude and direction).
Final momentum = (1.7 kg) x v' + (2.39 kg) x 0
Since momentum is conserved, we can set the initial and final momentum equal to each other:
-9.044 kg·m/s = (1.7 kg) x v'
Solving for v':
v' = -9.044 kg·m/s / 1.7 kg ≈ -5.32 m/s
Therefore, the velocity of the ball just after the collision is approximately 5.32 m/s upwards (opposite to the initial direction).