A force F = 20 N acts on a body of mass m = 200 g for Delta…

A force $F = 20\ \mathrm{N}$ acts on a body of mass $m = 200\ \mathrm{g}$ for $\Delta t = 0.03\ \mathrm{s}$. What is the impulse of the force, and what is the body's speed after the force stops, if it started from rest?

Explanation of Method

Impulse of a short push and the resulting speed.

Step 1: Impulse

\[ J = F\,\Delta t = 20 \cdot 0.03. \]

\[J = 0.6\ \mathrm{N\,s}\]

Step 2: Final speed

The impulse equals the change of momentum; the body starts at rest: \[J = m v - 0 \;\Longrightarrow\; v = \frac{J}{m} = \frac{0.6}{0.2}.\]

\[v = 3\ \mathrm{m/s}\]

Check

$ a = F/m = 100\ \mathrm{m/s^2} $, $ v = a\,\Delta t = 100 \cdot 0.03 = 3\ \mathrm{m/s}\ \checkmark $

$F\ (\mathrm{N})$ $t\ (\mathrm{s})$ $0$ $0.03$ $0.05$ $20$ $J = 0.6\ \mathrm{N\,s}$ $\text{then } v = \tfrac{J}{m} = 3\ \mathrm{m/s}$

Force against time. The impulse is the area under the graph, $20\ \mathrm{N} \times 0.03\ \mathrm{s}$; it all goes into the body's momentum.

Result

\[\boxed{\,J = 0.6\ \mathrm{N\,s}\,}\]

\[\boxed{\,v = 3\ \mathrm{m/s}\,}\]