Question

A force towards the -x axis is applied for 27ms to a 400g ball initially moving at 14m/s in the +x axis. The force varies in magnitude, and the impulse has a magnitude of 32.4kg m/s. What is the balls speed and direction of travel just after the force is applied?

Answer 1

Initial momentum:

$$p_i=mv_i=0.4\cdot14=5.6\text{ kg m/s}.$$

Final impulse will help us with finding the final speed:

$$-F\Delta t=m\Delta v,\\ -F\Delta t=m(v_f-v_i)=32.4\text{ kg}\cdot\text{m/s}.\\\space\\ v_f=v_i-\frac{F\Delta t}{m}=14-\frac{32.4\cdot0.027}{0.4}=11.8\text{ m/s}.$$