(1)
A large swinging balls and alls drive a horizontal iron spike into a vertical wall The center of the position srough a vertical height of 1.6 m bertical striking the spike in the shown.
wall
1.6m
ball mass 3.5 kg
spike mass
0.80 kg
The mass of the ball is 3.5 kg and the mass of the spike is 0.80 kg. Immediately after striking the spike, the ball and spike move together Find the
10 speed of the ball on striking the spike.
(ii) energy dissipated as a result of the collision.
(i) To find the speed of the ball on striking the spike, we can use the principle of conservation of momentum.
Let the initial velocity of the ball be v1 and the initial velocity of the spike be 0 (as it was at rest). After the collision, the two will move together with a common velocity v.
The total momentum before the collision is given by:
3.5kg * v1 = (3.5kg + 0.80kg) * v
Solving for v1, we get:
v1 = (3.5 + 0.80) * v / 3.5
v1 = 4.3v / 3.5
v1 = 1.23v
Given that the ball falls through a height of 1.6m, we can use the equation of energy conservation to find the final velocity:
mgh = 1/2 mv^2
(3.5kg * 9.81m/s^2 * 1.6m) = 1/2 * (3.5kg + 0.80kg) * v^2
55.44J = 1/2 * 4.3kg * v^2
v^2 = 25.8
v = 5.08 m/s
Therefore, the speed of the ball on striking the spike is 5.08 m/s.
(ii) The energy dissipated as a result of the collision can be found by calculating the initial kinetic energy of the ball and spike and subtracting the final kinetic energy after the collision.
Initial kinetic energy = 1/2 * 3.5kg * (1.23v)^2 + 1/2 * 0.80kg * 0^2
Final kinetic energy = 1/2 * (3.5kg + 0.80kg) * 5.08^2
Energy dissipated = Initial kinetic energy - Final kinetic energy
= [1/2 * 3.5 * (1.23 * 5.08)^2] - [1/2 * 4.3 * 5.08^2]
= 80.4 J
Therefore, the energy dissipated as a result of the collision is 80.4 Joules.