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5.Work, Energy, Power and Collision
medium
A body of mass $5 \,kg$ is moving with a momentum of $10\, kg-m/s$. A force of $0.2 \,N$ acts on it in the direction of motion of the body for $10\, seconds$. The increase in its kinetic energy is ............... $J$
A$2.8$
B$3.2$
C$3.8$
D$4.4$
Solution
(d)Change in momentum = Force $×$ time
${P_2} – {P_1} = F \times t = 0.2 \times 10 = 2$
$⇒$ ${P_2} = 2 + {P_1} = 2 + 10 = 12kg{\rm{ – }}m/s$
Increase in K.E. = $\frac{1}{{2m}}(P_2^2 – P_1^2)\;$
$= \frac{1}{{2 \times 5}}\left[ {{{(12)}^2} – {{(10)}^2}} \right]$
$ = \frac{{44}}{{10}} = 4.4\,J$
${P_2} – {P_1} = F \times t = 0.2 \times 10 = 2$
$⇒$ ${P_2} = 2 + {P_1} = 2 + 10 = 12kg{\rm{ – }}m/s$
Increase in K.E. = $\frac{1}{{2m}}(P_2^2 – P_1^2)\;$
$= \frac{1}{{2 \times 5}}\left[ {{{(12)}^2} – {{(10)}^2}} \right]$
$ = \frac{{44}}{{10}} = 4.4\,J$
Standard 11
Physics
