Answer
a) The speed of the 1.5-kg ball just before impact is $5.56m/s$
b) The velocity of the 1.5-kg ball after impact is $-2.82m/s$ and that of the 4.6-kg ball is $+2.73m/s$, assuming the direction of motion of the 1.5-kg ball before impact is the positive direction.
c) The 1.5-kg ball reaches $0.41m$ in height and the 4.6-kg ball reaches $0.38m$ in height.
Work Step by Step
a) Following the principle of conservation of mechanical energy, we have $$\frac{1}{2}m(v_f^2-v_0^2)+mg(h_f-h_0)=0$$ $$\frac{1}{2}(v_f^2-v_0^2)+g(h_f-h_0)=0 (1)$$
We have $v_0=5m/s$, $g=9.8m/s^2$ and $h_f-h_0=-0.3m$, so the speed of the 1.5-kg ball just before impact is $$v_f=5.56m/s$$
b) Assume the direction of motion of the 1.5-kg ball before impact is the positive direction.
Using the principle of conservation of total linear momentum, we have $$M\vec{V_0}+m\vec{v_0}=M\vec{V_f}+m\vec{v_f}$$ $$4.6\vec{V_f}+1.5\vec{v_f}=4.6\times0+1.5\times(+5.56)=8.34$$ $$3.07\vec{V_f}+\vec{v_f}=5.56 (2)$$
Since the collision is elastic, the kinetic energies are conserved, too: $$\frac{1}{2}(MV_0^2+mv_0^2)=\frac{1}{2}(MV_f^2+mv_f^2)$$ $$MV_0^2+mv_0^2=MV_f^2+mv_f^2$$ $$4.6V_f^2+1.5v_f^2=0+1.5(5.56)^2$$ $$3.07V_f^2+v_f^2=(5.56)^2=30.91 (3)$$
Solving (2) and (3), we get $\vec{V}_f=+2.73m/s$ and $\vec{v_f}=-2.82m/s$
c) We again apply equation (1) to find how high each ball swings:
- The 1.5-kg ball: $v_0=2.82m/s$, $v_f=0$, $g=9.8m/s^2$ $$\frac{1}{2}(-2.82^2)+(9.8)\Delta h=0$$ $$\Delta h=0.41m$$
- The 4.6-kg ball: $v_0=2.73m/s$, $v_f=0$, $g=9.8m/s^2$ $$\frac{1}{2}(-2.73^2)+(9.8)\Delta h=0$$ $$\Delta h=0.38m$$
