MCQ
A force $\vec{F}=4 t^3$ is applied on an object for the first two seconds. The linear momentum of object will increase.
- ✓$16 N- s$
- B$8 N- s$
- C$48 N- s$
- D$32 N- s$
✓
Answer
Correct option: A.
$16 N- s$
(A) Impulse acting on object :
$
\begin{aligned}
J & =\int_{t_1}^{t_2} F d t=\int_0^2 4 t^3 d t \\
J & =4\left[\frac{t^4}{4}\right]_0^2=\left[t^4\right]_0^2=2^4-0 \\
& =16 N-s
\end{aligned}
$
From the impulse-momentum theorem,
change in linear momentum $=$ Impulse $=16 Ns$
Hence, correct option is (A).
$
\begin{aligned}
J & =\int_{t_1}^{t_2} F d t=\int_0^2 4 t^3 d t \\
J & =4\left[\frac{t^4}{4}\right]_0^2=\left[t^4\right]_0^2=2^4-0 \\
& =16 N-s
\end{aligned}
$
From the impulse-momentum theorem,
change in linear momentum $=$ Impulse $=16 Ns$
Hence, correct option is (A).
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