Question

Two pendulum bobs of mass m and 2m collide elastically at the lowest point in their motion. If both the balls are released from height h above the lowest point velocity of the bob of mass m just after collision​

Answer 1

The velocity of bob after collision is V1 = 5/3√2gH

Explanation:

From energy conservation speed with which both the balls strike

mgh = 1/2 mVo^2

Vo  = √2gh

Velocity of m after collision

V1 = (m1−m2/m1+m2) u1+2m2/m1+m2u2

V1=(m−m×2/m+2m)(−√2gH)+(4m/m+2m)√2gH

V1 = 5/3√2gH

The velocity of bob after collision is V1 = 5/3√2gH

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Answer 2

The velocity at the lowest point of the bob after collision is $\dfrac{5}{3}\sqrt{2gh}$

Explanation:

Given that,

Mass of first bob = m

Mass of second bob = 2m

Height = h

We need to calculate the velocity of the bob after collision

Using formula of collision

$v_{1}=\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}}u_{1}+\dfrac{2m_{2}}{m_{1}+m_{2}}u_{2}$

Put the value into the formula

$v_{1}=\dfrac{m-2m}{m+2m}\times(-\sqrt{2gh})+\dfrac{4m}{m+2m}\times\sqrt{2gh}$

$v_{1}=\dfrac{\sqrt{2gh}}{3}+\dfrac{4\sqrt{2gh}}{3}$

$v_{1}=\dfrac{5}{3}\sqrt{2gh}$

Hence, The velocity at the lowest point of the bob after collision is $\dfrac{5}{3}\sqrt{2gh}$

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