A ball of mass m moving with velocity v strikes the bob of a pendulum at rest. The mass of the bob is also m . If the collision is perfectly…

Question

A ball of mass m moving with velocity v strikes the bob of a pendulum at rest. The mass of the bob is also m . If the collision is perfectly inelastic, the height to which the bob will rise is given by h =(v^2)/x.g , then x=?

MARKS Content Team · Accepted Answer

Correct answer: 8. Because the collision is perfectly inelastic, hence the two blocks stick together. By conservation of linear momentum, 2 mV = mv or V = v/2 By conservation of energy, 2mgh=1/2 2mV 2^2 or h = v^2 / 8 g x=8

A ball of mass $m$ moving with velocity $v$ strikes the bob of a pendulum at rest. The mass of the bob is also $m$ . If the collision is perfectly inelastic, the height to which the bob will rise is given by $h = \frac{v^{2}}{x .g},$ then $x = ?$

Solution

More Center of Mass Momentum and Collision Questions

A ball of mass m moving with velocity v strikes the bob of a pendulum at rest. The mass of the bob is also m . If the collision is perfectly inelastic, the height to which the bob will rise is given by h = v 2 x.g , then x=?

Solution

More Center of Mass Momentum and Collision Questions

A ball of mass $m$ moving with velocity $v$ strikes the bob of a pendulum at rest. The mass of the bob is also $m$ . If the collision is perfectly inelastic, the height to which the bob will rise is given by $h = \frac{v^{2}}{x .g},$ then $x = ?$