A 0.160-kg hockey puck is moving on an icy, frictionless, horizontal surface. At t = 0, the puck is moving to the right at 3.00 m/s. (a) Calculate the velocity of the puck (magnitude and direction) after a force of 25.0 N directed to the right has been applied for 0.050 s. (b) If, instead, a force of 12.0 N directed to the left is applied from t = 0 to t = 0.050 s, what is the final velocity of the puck?

Solution 9E Step 1: a) Force acting on the hockey puck to the right, F = 25 N Time, t = 0.05 seconds We know that, the force, F = ma Where, m - mass of the hockey puck and a - acceleration of the hockey puck a = (v - u) / t Where, v - final velocity of the puck and u - initial velocity of the puck Provided, u = 3 m/s and m = 0.160 kg So, we can write, 25 N = 0.160 kg (v - 3 m/s) / 0.05 s 0.160 kg (v - 3 m/s) = 25 N × 0.05 s 0.160 kg v - 0.480 kg m/s = 1.25 Ns 2 0.160 kg v = 0.480 kg m/s + 1.25 kg m/s ( We know that, Ns = kgm/s × s = kg m/s) 0.160 kg v = 1.73 kg m/s v = 1.73 kg m/s / 0.160 kg v = 10.81 m/s