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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.
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: \[F = \frac{p}{\Delta t} = \frac{25\ \mathrm{Ns}}{12s} = 2.08\ \mathrm{N}\]