impulse =change of momentum
as in
F = d/dt (m V) = dmV / dt which if m is constant = m A
so F dt = d (m V) = change of momentum
so here
original m v = + .23 * 20
final momentum = - .23 * 15
change = final - original =.23 (-15 - 20) = - .23 * 35 = - 8.05 kg m/s
A basketball of mass 0.23 kg is thrown horizontally against a rigid vertical wall with a velocity of 20 m/s. It rebounds with a velocity of 15 m/s. Calculate the impulse of the force of the wall on the basketball.
1 answer