Question

Two identical masses A and B are on same vertical line. A is released from rest from a height while B is simultaneously projected upwards with a velocity of 50 m/s. Find the velocity of the combined mass just after collision considering perfectly inelastic
collision. (g = 10 m/s2)
(1) 20 m/s
(3) 15 m/s
(2) 5 m/s
(4) 25 m/s​

Answer 1

Answer:

Given:

Masses are same. Height of vertical line is 50 m.

To find:

Velocity of combined mass ( considering perfectly in-elastic collision.

Calculation:

Let ball A be dropped from top and ball B be projected from down.

Let the balls meet at "x" distance from top .

For Ball A :

$\sf{x = \dfrac{1}{2} g {t}^{2} }$

For Ball B :

$\sf{(50 - x )= 50t - \dfrac{1}{2} g {t}^{2} }$

Adding the 2 Equations :

$\sf{ \therefore \: 50 = 50t}$

$\sf{ \implies \: t = 1 \: sec}$

So speed of Ball A just before Collision :

$\sf{ \therefore \: v_{A} = 0 + gt}$

$\sf{ \implies \: v_{A} = 10 \times 1 = 10 \: m {s}^{ - 1} }$

Similarly, speed of Ball B just before Collision :

$\sf{ \therefore \: v_{B} = u - gt}$

$\sf{ \implies \: v_{B} = 50 - (10 \times 1)}$

$\sf{ \implies \: v_{B} = 40 \: m {s}^{ - 1} }$

Finally applying Conservation of Momentum :

$\sf{m_{A}v_{A} - m_{B}v_{B} = (m_{A} + m_{B})v }$

$\sf{ \implies \: m(40) - m(10) = 2m(v)}$

$\sf{ \implies \: 2m(v) = m(30)}$

$\sf{ \implies \: v = 15 \: m {s}^{ - 1}}$

So final answer :

$\boxed{ \huge{ \red{ \bold{\sf{ \: v = 15 \: m {s}^{ - 1}}}}}}$

Answer 2

Explanation:

follow the attachment.

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