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 2.5 kg, and the length of the wire is 1.19 m. Find the velocity (magnitude and direction) of the ball (a) just before the collision, and (b) just after the collision.

Question

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 2.5 kg, and the length of the wire is 1.19 m. Find the velocity (magnitude and direction) of the ball (a) just before the collision, and (b) just after the collision.

Answer 1

To find the velocity (magnitude and direction) of the ball just before the collision, we can use energy conservation.

Let's denote the velocity of the ball just before the collision as Vb, and the velocity of the block just before the collision as Vbl.

1. We will start by finding the potential energy of the ball when it is released from rest:
Potential Energy (PE) = m * g * h
PE = 1.7 kg * 9.8 m/s^2 * 1.19 m
PE = 19.1454 J

2. At the lowest point, where the ball is about to collide with the block, all the potential energy is converted into kinetic energy. Therefore, the kinetic energy just before the collision is:
Kinetic Energy (KE) = PE = 19.1454 J

3. The kinetic energy of the ball is given by:
KE = (1/2) * m * Vb^2
19.1454 J = (1/2) * 1.7 kg * Vb^2
Vb^2 = (2 * 19.1454 J) / 1.7 kg
Vb^2 = 22.6 m^2/s^2
Vb = √(22.6) m/s
Vb ≈ 4.75 m/s

So, the magnitude of the velocity of the ball just before the collision is approximately 4.75 m/s (rounded to two decimal places).

Now let's find the velocity of the ball just after the collision:

(a) To find the velocity of the block just before the collision, we can use the conservation of momentum. The momentum of the system is conserved because there are no external horizontal forces acting.
Momentum before the collision = Momentum after the collision
(m * Vb) + (m * Vbl) = (m * Vb') + (m * Vbl')
Since the ball comes from rest, Vbl is zero, and the equation simplifies to:
1.7 kg * Vb = 1.7 kg * Vb' + 2.5 kg * Vbl'

(b) Since the collision is elastic, the relative velocity of separation will be equal to the relative velocity of approach. Therefore,
Vb' - Vbl' = -(Vb - Vbl)

Now let's substitute the given values:
1.7 kg * 4.75 m/s = 1.7 kg * Vb' + 2.5 kg * 0
8.075 kg·m/s = 1.7 kg * Vb'
Vb' = 8.075 kg·m/s / 1.7 kg
Vb' ≈ 4.75 m/s

Therefore, the magnitude of the velocity of the ball just after the collision is approximately 4.75 m/s (rounded to two decimal places). The direction is the same as the initial velocity before the collision.