0
$\begingroup$

The motion of a particle of mass $m$ is given by $x = 0$ for $t < 0$, $x(t) = A \sin 4pi t$ for $0 < t <(1/4) s$ ($A > 0$), and $x = 0$ for $t >(1/4) s$.

Which of the following statements is true?

(a) The force at t=(1/8)s on the particle is –16Amπ^2.

(b) The particle is acted upon by on impulse of magnitude 4Amπ^2 at t=0 s and t = (1/4)s.

(c) The particle is not acted upon by any force.

(d) The particle is not acted upon by a constant force.

(e) There is no impulse acting on the particle.

MY THOUGHTS

(a) is easy. I took double derivative of x(t) to get a(t). Then found F(t)= m * a(t).

(c) is dead wrong obviously. And (d) is correct as force is time varying.

(e) should also be wrong as momentum will change due to change in velocity with time.

Now my confusion is for (b).

Now as far as I know, impulse can be calculated in 2 ways-

(i) Change in momentum

(ii) integral of force over time interval

I don't know how to find change in momentum directly neither integral for a particular time instant.

Tl;dr How to find impulse at a particular time instant.

$\endgroup$

1 Answer 1

0
$\begingroup$

The velocity of the particle at time $t$ is $\frac{dx(t)}{dt}$ and so its momentum is $m\frac{dx(t)}{dt}$. Look at how $\frac{dx(t)}{dt}$ behaves close to $t=0$ and close to $t=\frac 1 4$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.