A 200 g particle gains a momentum of 25 Ns in 12 s starting from rest. Find the
kinetic energy of the particle and the applied force.

Question

A 200 g particle gains a momentum of 25 Ns in 12 s starting from rest. Find the
kinetic energy of the particle and the applied force.

anonymous · Answer

We are given the mass of the particle m=200 g = 0.2 kg.
And the momentum p = 25 Ns = 25 kg m/s.
We know $p=mv$ and kinetic energy $E_k = \frac{1}{2}mv^2$.
Multiplying numerator and denominator of the expression for $E_k$ by m, we get:
$$E_k = \frac{p^2}{2m} $$
Plugging in our given quantities, yields:
$$E_k = 1562.5\ \mathrm{J}$$
Assuming a constant force, we can use the impulse/momentum equation:
$$F\Delta t = \Delta p$$
Since we are starting from rest $\Delta p = p$, and we are given $\Delta t = 12\mathrm{s}$.
So that:
$$F = \frac{p}{\Delta t} = \frac{25\ \mathrm{Ns}}{12s} = 2.08\ \mathrm{N}$$