Q.A body of mass $0.5\ \text{kg}$ moving in a straight line has its velocity changed from $5\ \text{m/s}$ to $15\ \text{m/s}$ (in the same direction) in $2\ \text{s}$ by a constant force. Find the magnitude of the force and the impulse delivered to the body.

Given: mass $m = 0.5\ \text{kg}$, initial velocity $u = 5\ \text{m/s}$, final velocity $v = 15\ \text{m/s}$ (same direction), time $t = 2\ \text{s}$.

Force: the acceleration is $a = \frac{v - u}{t} = \frac{15 - 5}{2} = 5\ \text{m/s}^2$ so by Newton's second law, $F = ma = 0.5 \times 5 = 2.5\ \text{N}$

Impulse: using the impulse-momentum theorem, $J = F\,t = 2.5 \times 2 = 5\ \text{kg}\,\text{m/s}$ This can be checked directly from the change in momentum: $\Delta p = m(v-u) = 0.5 \times (15-5) = 0.5 \times 10 = 5\ \text{kg}\,\text{m/s}$, exactly matching the impulse found above, as the impulse-momentum theorem requires.

Find acceleration from $a = (v-u)/t$, then $F = ma$; find impulse either as $F\,t$ or directly as $\Delta p = m(v-u)$ -- both must agree.

• Computing impulse and then treating it as a separate, unrelated quantity from the change in momentum, rather than checking the two values agree.
• Forgetting to divide the velocity change by the correct time interval when finding acceleration.