Question

If a body of mass 5g initially at rest is acted upon by a force of 50 dynes for 3 s. Then the impulse will be
(A) $0.98 \times {10^{ - 3}}Ns$
(B) $1.5 \times {10^{ - 3}}Ns$
(C) $2.0 \times {10^{ - 3}}Ns$
(D) $2.5 \times {10^{ - 3}}Ns$

Answer 1

Hint: We know that impulse is the product of force acting and time of contact, substitute these values in the formula $I = F\Delta t$ to obtain the impulse.

Complete step-by-step solution
When a large force acts on a body for a very small-time interval it is called impulsive force. Impulse of a force is a measure of total effect of force.
It is given by,
 $I = F\Delta t$
Here, F = force and \[\Delta t\] is the time of contact.
It is given that \[F = 50\] dynes
 $
  1dyne = {10^{ - 5}}N \\
  50dynes = 50 \times {10^{ - 5}}N \\
 $
 $\Delta t = 3s$
Substituting the known data in the above expression we get,
 $
  I = 50 \times {10^{ - 5}} \times 3 \\
  I = 1.5 \times {10^{ - 3}}Ns \\
 $

Hence, the impulse will be \[1.5 \times {10^{ - 3}}\] Ns and the correct option is B.

Note: An impulsive force does not remain constant but changes from zero to maximum and then from maximum to zero. It is a vector quantity and its direction is the same as that of force. From the impulse momentum theorem impulse of a force is equal to the change in momentum.
 $I = \Delta p$