Question

A $10kg$ wagon is pushed with a force of $7N$ for $1.5second$, then with a force of $5N$ for $1.7seconds$, and then with a force of $10N$ for $3$ second in the same direction. What is the change in velocity brought about?

• A. $9.8m/s$
• B. $19.6m/s$
• C. $4.9m/s$
• D. $10m/s$

Answer 1

The correct option is C $4.9m/s$

Given that,

Mass of wagon $m=10kg$

Force applied $F=7N$

Time duration $t=1.5s$

Initial velocity $=v_1$

Final velocity $=v_2$

By using impulse -change in momentum equation

$m(v_2-v_1)=Ft$

$10(v_2-v_1)=7\times 1.5$

$v_2-v_1=1.05$

$v_2=1.05+v_1......\left(I\right)$

Now, force and time is

Force $F=5N$

Time $t=1.7s$

Initial velocity $=v_2$

Final velocity$=v_3$

By using impulse -change in momentum equation

$m(v_3-v_2)=Ft$

$10(v_3-v_1-1.05)=5\times 1.7$

$v_3=v_1+\mathrm{1.9....}\left(II\right)$

Now,

Force $F=10N$

Time $t=3s$

Initial velocity $=v_3$

Final velocity $=v_4$

By using impulse -change in momentum equation

$m(v_4-v_3)=Ft$

$v_4-v_1-1.9=3$

$v_4-v_1=4.9m/s$

Hence, the change in velocity is $4.9m/s$