GATE Mechanical Engineering (ME) 2018 Shift 1 Solved Paper

Question : 64 of 65

Marks: +1, -0

A point mass is shot vertically up from ground level with a velocity of

4 {m} {/} {s}

at time,

t=0 .

It loses

20 \%

of its impact velocity after each collision with the ground. Assuming that the acceleration due to gravity is

10 {m} {/} {s}^{2}

and that air resistance is negligible, the mass stops bouncing and comes to complete rest on the ground after a total time (in seconds) of

Solution:

(1) →

t \;\; =?

v \;\; =u+a t

0 \;\; =4-10 t

t=\;{4}/{10}=0.4 s

(2) →

t{'} \;\; =?

u{'} \;\; =0.8 × u

\;\; =0.8 × 4=3.2 {m} {/} {s}

v{'} \;\; =u{'}+a t{'}

0 \;\; =3.2-10 t{'}

t{'}=\;{3.2}/{10}=0.32 {s}

3 →

t{'} \;\; =?

u{'} \;\; =0.8 u{'}

\;\; =0.8 × 3.2=2.56 {m} {/} {s}

v{''} \;\; =u{''}+a t{''}

0 \;\; =2.56-10 t{''}

t{''} \;\; =0.256 s

So,

t, t{'}, t{''}

are forming a GP series
So, total time

\;\; =2(t+t{'}+t{''}+. . . . . . .0)

\;\; =2[0.4+0.32+0.256+. . . .0]

\;\; =2 × \;{0.4}/{1-0.8}=2 × 2=4 {s}

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