Question

A block of mass $m=3kg$ resting on a horizontal frictionless floor is horizontally struck by a $9N$ force that acts for $0.02$ sec. After $3$ sec it receives a second blow of force 9N but in an opposite direction which acts for $0.01$ sec, the speed of the body after $30$ sec is:

• A. $0$
• B. $\text{3 cm/sec}$
• C. $\text{90 cm/sec}$
• D. $\text{30 cm / sec}$

Answer 1

The correct option is B $\text{3 cm/sec}$

The mass initially at rest (m = 3 kg) in struck by a force $F_1=9N$ for $t_1=0.025$

$\stackrel{F_1}{\to }\stackrel{V_1=0m/s}{m}\Rightarrow \underset{m}{\overset{V_2}{\to }}$

Due to the impulse given by force $F_1$, mass attains velocity $V_2$

As impulse :

$J=F\Delta t=m\Delta v$

$\Rightarrow 9\times 0.02=m(V_2-V_1)[V_1=0]$

$\Rightarrow 3\times V_2=0.18\Rightarrow V_2=0.06m/s$

Again another force $F_2=9N$ acts for $t_2=0.01s$ in opposite direction.

$\underset{m}{\overset{V_1}{\to }}\leftarrow F_2\Rightarrow {\underset{m}{\overset{V}{\to }}}_3$

Due to impulse given by $F_2$, mass finally attain velocity $V_3$.

Impulse :

$J=F_2{t}_2=-m\Delta v$ (as $F_2$ acts in opposite direction)

$\Rightarrow 9\times 0.01=-3\times (V_3-V_2)$

$\Rightarrow V_2-V_3=0.03$

$\Rightarrow V_3=0.06-0.03=0.03m/s$

$V_3=3$ cm/s