GATE Mechanical Engineering (ME) 2020 Shift 2 Solved Paper

Question : 46 of 65

Marks: +1, -0

A rigid block of mass

m_{1}=10 {kg}

having velocity

v_{0}=2 {m} {/} {s}

strikes a stationary block of mass

m_{2}=30 {kg}

after travelling

1 {m}

along a frictionless horizontal surface as shown in the figure.

The two masses stick together and jointly move by a distance of

0.25 {m}

further along the same frictionless surface, before they touch the mass-less buffer that is connected to the rigid vertical wall by means of a linear spring having a spring constant

{k}=10^{5} {N}{/} {m}

. The maximum deflection of the spring is

{cm}

(round off to 2 decimal places).

Your Answer:

Validate

Solution:

{m}_{1}=10 {kg}

{v}_{0}=2 {m} {/} {s}

{m}_{2}=30 {kg}

Since given surfaces are friction less surface. Thus, both energy and linear momentum will be conserve.

{k}=10^{5} {N} {/} {m}

After collision both mass sticks together. Thus,

10 × 2+30 × 0=(10+30) V

common

20=40 {V}_{\;\text "common "\;}

{V}_{\;\text "common "\;}=\;{1}/{2} {m} {/} {s}

\;{1}/{2}( {m}_{1}+ {m}_{2}) {V}_{\;\text "common "\;}^{2}=\;{1}/{2} {kx}^{2}

40 ×(\;{1}/{2})^{2}=10^{5} × {x}^{2}

10 × 10^{-5}= {x}^{2}

{x}=10^{-2} {m}

{x}=1 {cm}

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