Answer to Question #157717 in Physics for Andu

Question

Two spheres with masses m 1 = 800 g and m2 = 400 g move along a line following each other with velocities v1 = 5 m / s and v2 = 4m / s. a) Calculate the pulse of the system. b) Will the spheres be hit? Why ? c) What will be the impulse of the system after the shock?

Answer 1

(a) By the definition of the impulse, we have:

$$J=F\Delta t=\Delta p=m\Delta v.$$

Let's find the initial momentum of the system of spheres before the collision:

$$p_i=m_1v_{1i}+m_2v_{2i},$$ $$p_i=0.8\ kg\cdot5\ \dfrac{m}{s}+0.4\cdot4\dfrac{m}{s}=5.6\ \dfrac{kgm}{s}.$$

Therefore, the impulse of the system before the collision equals $5.6\ \dfrac{kgm}{s}$.

(b) Since the first sphere has the greater velocity than the second one, the both spheres will collide with each other.

(c) Let's find the final velocities of the spheres after the collision from the law of conservation of momentum:

$$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}.$$

Let's also assume that collision is perfectly elastic, therefore, the kinetic energy will conserved:

$$\dfrac{1}{2}m_1v_{1i}^2+\dfrac{1}{2}m_2v_{2i}^2=\dfrac{1}{2}m_1v_{1f}^2+\dfrac{1}{2}m_2v_{2f}^2.$$

Solving this two equations we can find the final velocities of the spheres after the collision:

$$v_{1f}=\dfrac{(m_1-m_2)v_1+2m_2v_2}{(m_1+m_2)},$$ $$v_{1f}=\dfrac{(0.8\ kg-0.4\ kg)\cdot5\ \dfrac{m}{s} +2\cdot0.4\ kg\cdot4\ \dfrac{m}{s}}{(0.8\ kg+0.4\ kg)}=4.3\ \dfrac{m}{s},$$ $$v_{2f}=\dfrac{2m_1v_1+(m_2-m_1)v_2}{(m_1+m_2)},$$ $$v_{2f}=\dfrac{2\cdot0.8\ kg\cdot5\ \dfrac{m}{s}+(0.4\ kg-0.8\ kg)\cdot4\ \dfrac{m}{s}}{(0.8\ kg+0.4\ kg)}=5.3\ \dfrac{m}{s}.$$

Finally, we can find the final momentum of the system of spheres after the collision:

$$p_f=m_1v_{1f}+m_2v_{2f},$$ $$p_f=0.8\ kg\cdot4.3\ \dfrac{m}{s}+0.4\cdot5.3\dfrac{m}{s}=5.6\ \dfrac{kgm}{s}.$$

Therefore, the impulse of the system after the collision equals $5.6\ \dfrac{kgm}{s}.$