Question

25. A stone weighing 200g is thrown up with an initial velocity of V = 6 m / s. a) What happens to the impulse during the ascent, increases or decreases? Why? b) Calculate the change in momentum as the stone reaches its greatest height.

Answer 1

a) Impulse is the change in momentum, so we can write:

$$J=F\Delta t=\Delta p=m(v_f-v_i),$$ $$J=m(v_f-v_i)=-F\Delta t.$$

Therefore, the impulse will decreases.

b)

$$\Delta p=m(v_f-v_i)=0.2\ kg\cdot(0-6\ \dfrac{m}{s})=-1.2\ \dfrac{kgm}{s}.$$