Gujarati
Hindi
5.Work, Energy, Power and Collision
normal

A particle of mass $7\, kg$ moving at $5\, m/s$ is acted upon by a variable force opposite to its initial direction of motion. The variation of force $F$ is shown as a function of time $t$.

A

at $t = 10\, s$ speed of particle is $5\, m/s$

B

at $t = 10\, s$ direction of motion of particle is reversed

C

at $t = 5\, s$ the particle will momentarily at rest

D

all the above options are correct.

Solution

Initial momentum $=35 \mathrm{\,Ns}$

Change in momentum $=$ area of graph

$=\frac{14 \times 10}{2}=70 \mathrm{\,Ns}$

Final momentum $=-35 \mathrm{\,Ns}$

Standard 11
Physics

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