Answer to Question #117411 in Quantum Mechanics for Mehmet Fazıl Kıymazaslan

Question

A pendulum is made byattaching a ball of mass m=0.2kg to the end of a massless string. The ball is swung out until it is at a height h=0.2 m above the ground and then is released from rest. When the ball is at the lowest point in the swing, it collides with a ball of mass M=0.4kg originally at rest and sticksto it. The combined masses swing together to the right and rise to a maximum height H. Assume that there is negligible friction.
(a) What is the speed of the ball with mass m just before it strikes the second ball resting on the ground?
(b) What is the speed of the combined masses right after the collision?
(c) Calculatethe maximum height H reached by the combined masses.

Answer 1

a) Let us write the law of conservation of energy for the first body

$mgh = \dfrac{mv^2}{2} \Rightarrow v = \sqrt{2gh}= \sqrt{2\cdot 10\cdot0.2} = 2\,\mathrm{m/s}.$

b) According to the law of conservation of momentum

$mv + M\cdot0 = (m+M)V.$ Therefore, $V = \dfrac{mv}{M+m} = \dfrac{0.2\cdot2}{0.4+0.2} \approx 0.67\,\mathrm{m/s}.$

c) Let us write the law of conservation of energy for the two bodies:

$(m+M)gH = \dfrac{(m+M)V^2}{2} \Rightarrow H = \dfrac{V^2}{2g} \approx 0.022\,\mathrm{m}.$