GATE Mechanical Engineering (ME) 2019 Shift 2 Solved Paper

Question : 53 of 65

Marks: +1, -0

A ball of mass

3 {kg}

moving with a velocity of

4 {m} {/} {s}

undergoes a perfectly-elastic direct-central impact with a stationary ball of mass

m

. After the impact is over, the kinetic energy of the

3 {kg}

ball is

6 {J}

. The possible value (s) of

{m}

is/are

Solution:

Conservation of momentum gives

m_{1} u_{1}+m_{2} u_{2}=m_{1} v_{1}+m_{2} v_{2}

m_{1}=3 {kg}, u_{1}=4 {m}{/} {s}, m_{2}=m {kg}, u_{2}=0

Substituting in the above equation,

3 × 4+m × 0=3 × v_{1}+m × v_{2}

Or,

12=3 v_{1}+m v_{2}

.....(1)
For perfectly-elastic impact,

e=\;{v_{2}-v_{1}}/{u_{1}-u_{2}}=1

Or,

v_{2}-v_{1}=u_{1}-u_{2}=4 {m}{/} {s}

......(2)
Conservation of kinetic energy gives

\;{1}/{2} m_{1} u_{1}^{2}+\;{1}/{2} m_{2} u_{2}^{2}=\;{1}/{2} m_{1} v_{1}^{2}+\;{1}/{2} m_{2} v_{2}^{2}

Or,

\;{1}/{2} 3(4)^{2}+0=6+\;{1}/{2} m_{2} v_{2}^{2}

Or,

m v_{2}^{2}=36

.............. (3)
From equation (2),

v_{1}=v_{2}-4

Substituting in equation (1),

12=3(v_{2}-4)+m v_{2}

12=3 v_{2}-12+m v_{2}

v_{2}(3+m)=24

v_{2}=\;{24}/{3+m}

Substituting in equation (3)

m(\;{24}/{3+m})^{2}=36

576 m=36(3+m)^{2}

⇒ 16 m=9+m^{2}+6 m

m^{2}-10 m+9=0

On solving, we get
𝑚 = 1,9
Therefore, possible values of m are 1 kg and 9 kg.

Go to Question:

Prev Question

Next Question