Name: Work Done on a Block
Uploaded: 2019-07-19
Duration: 3 min 57 s
Description: Check out this solved illustration based on calculating work done on an object by forces acting at different angles. Just click this video to explore!

Question

Ball

A

, of mass

2 kg

, is moving in a straight line at

5 {ms}^{- 1}

. Ball

B

, of mass

4 kg

, is moving in the same straight line at

2 {ms}^{- 1}

. Ball

B

is travelling directly towards ball

A

. The balls hit each other and after the impact each ball has reversed its direction of travel. The kinetic energy lost in the impact is

12.5 J

.
Show that the speed of ball

A

after the impact is

\frac{10}{3} m s^{- 1}

.

Accepted Answer

..

Given,

Mass of Ball A, $m_{a} = 2 K g$

Initial Velocity of Ball A, $u_{a} = 5 m s^{- 1}$

Mass of Ball B, $m_{b} = 4 K g$

Initial Velocity of Ball B, $u_{b} = 2 m s^{- 1}$

To Show= $S p e e d o f b a l l A a f t e r i m p a c t = \frac{10}{3} m s^{- 1}$

We know that,

Kinetic Energy, $K . E . = \frac{1}{2} m v^{2}$

$∴ K . E . o f B a l l A = \frac{1}{2} \times 2 \times {(5)}^{2} = 25 J$

$K . E . o f B a l l B = \frac{1}{2} \times 4 \times {(2)}^{2} = 8 J$

$T o t a l K . E . (b e f o r e c o l l i s i o n) = K . E . o f B a l l A + K . E . o f B a l l B K . E . = 25 J + 8 J K . E . = 33 J$

Now, by Conservation of momentum,

Initial momentum=Final momentum i.e.

$m_{a} \times u_{a} + m_{b} \times u_{b} = m_{a} \times v_{a} + m_{b} \times v_{b}$

$\Rightarrow 2 \times 5 + 4 \times (- 2) = 2 \times (- v_{a}) + 4 \times v_{b} \Rightarrow 10 - 8 = - 2 v_{a} + 4 v_{b} \Rightarrow 1 = - v_{a} + 2 v_{b} (i)$ (since, velocity of ball B is in opposite direction)

Now,

$T o t a l K . E . (a f t e r c o l l i s i o n) = \frac{1}{2} \times m_{a} \times {v_{a}}^{2} + \frac{1}{2} \times m_{b} \times {v_{b}}^{2}$

= $= \frac{1}{2} \times 2 \times {v_{a}}^{2} + \frac{1}{2} \times 4 \times {v_{b}}^{2} = {v_{a}}^{2} + 2 {v_{b}}^{2} (i i)$

Equating Initial and Final K.E., we get

Total K.E. (before collision)=Total K.E. (after collision) $+ 12.5 J$ (According to Question)

$\Rightarrow 33 J = {v_{a}}^{2} + 2 {v_{b}}^{2} + 12.5 J$

$\Rightarrow {v_{a}}^{2} + 2 {v_{b}}^{2} = 20.5 J$ (iii)

Putting equation(i) in equation(iii), we get

${v_{a}}^{2} + 2 {\{\frac{(1 + v_{a})}{2}\}}^{2} = 20.5 {2 v_{a}}^{2} + 1 + {v_{a}}^{2} + 2 v_{a} = 41 3 {v_{a}}^{2} + 2 v_{a} - 40 = 0 3 {v_{a}}^{2} + 12 v_{a} - 10 v_{a} - 40 = 0 3 v_{a} (v_{a} + 4) - 10 (v_{a} + 4) = 0 (v_{a} + 4) (3 v_{a} - 10) = 0 v_{a} = \frac{10}{3} m s^{- 1}, - 4 m s^{- 1}$

Hence, proved

Important Questions on Work and Energy

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