The velocity of a body of mass $10\ kg$ increases from $4\ m/s$ to $8\ m/s$ when a force acts on it for $2\ s$.

(a) What is the momentum before the force acts?

(b) What is the momentum after the force acts?

(c) What is the gain in momentum per second?

(d) What is the value of the force?

 Here mass $m=10\ kg$

Initial velocity $u=4\ m/s$

Final velocity $v=8\ m/s$

Time $t=2\ s$

(a) Momentum before the force acts $P_1=mu=10\ kg\times4\ m/s=40\ kg.-m/s$

(b) Momentum after the force acts $P_2=mv=10\ kg\times8\ m/s=80\ kg.m/s$

(c) Gain in momentum $=P_2-P_1$

$=80\ kg-m/s-40\ kg-m/s$

$=40\ m/s$

Gain in momentum per second $=\frac{P_2-P_1}{t}$

$=\frac{40\ kg-m/s}{2\ s}$

$=20\ kg-m/s^2$

(d) As known force is equal to the rate of the change of the momentum.

Therefore, $F=\frac{P_2-P_1}{t}$

$=\frac{80kg-m/s-40kg-m/s}{2\ s}$

$=\frac{40\ kg-m/s}{2\ s}$

$=20\ kg-m/s^2=20/ N$

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Updated on: 10-Oct-2022

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