A S.00~kg steel ball strikes a wall with a speed of 12.0 m/s at an angle ofi9 = 60.0\" with the surface. It bounces off with the same speed and angle (see figure below). If the ball is in contact with the wall for 0.200 5, what is the average force exerted by the wall on the ball? (Assume right is the positive direction.) Part1 of 4 - Conceptualize If we think about the angle as a variable and consider the limiting cases, then the force should be zero when the angle is 0 (no contact between the ball and the wall). When the angle is 90 the force will be its maximum and can be found from the momentum-impulse equation as (5.00 kg)(2)(12.0 m/s) / 0.200 s = 600 N, the doubling coming from the reversal of velocity since there is no y component. For H = 60", we expect a force of less than this amount. Because of the symmetry about the normal to the surface, we expect the force must be directed to the left. Pan 2 of 4 - Categorize We use the momentumimpulse equation to find the force, and carefully consider the direction of the velocity vectors. We dene up as the positive y direction and to the right as the positive x direction. 3 of 4 - Analyze We use the following force-impulse equation. _. _, AP : PM = A(mv) The y component of momentum gives Apy =m(VY Viv) =rn(v cos 600 V cos 60.0) = 0, so the wall does not exert a force on the ball in the y direction. The x component gives AP = "7(fo Vix) = m ((l:l 'y )vsin 60.0 (l+_v| v )vsin 60.0\") = :l mv sin 60.0 and we have that Apx = :1 (C kg)( S m/s)(o-sss) = ll k9 . m/s