Question

5. Ball 1 has a mass of 2.0 kg and is suspended with a 3.0 m rope from a post so that the ball is stationary. Ball 2 has a mass of 4.0 kg and is tied to another rope. The second rope also measures 3.0 m but is held at a 60.0° angle, as shown in Figure 7. When ball 2 is released, it collides, head-on, with ball I in an elastic collision. (a) Calculate the speed of each ball immediately after the first collision. (b) Calculate the maximum height of each ball after the first collision. 3.0 m 60.0 3.0 m 4.0 kg 2.0 kg

Answer 1

DATA

$L=3 \, m$

$m_{1}=2 \, kg$

$m_{2}=4 \, kg$

$\theta =60 ^{\circ}$

$v_{1After} =?$

$v_{2After} =?$

$h_{1After} =?$

$h_{2After} =?$

Part (a) - Solution

$\:$

Since the collision is ellastic and the ball 1 is initially at rest, their final velocities (their velocities inmidietlly after collision) can be expressed as

$\:$

$v_{1After}=\frac{2m_{2}v_{2i}}{m_{1}+m_{2}} \: \: \: \: \: \: \: \: \: \: (1)$

$v_{2After}=\frac{(m_{2}-m_{1})v_{2i}}{m_{1}+m_{2}} \: \: \: \: \: \: \: \: \: \: (2)$

$\:$

We can use principle of conservation of mechanical energy to find initial velocity v_2i.

$\:$

$E_{top}=E_{bottom}$

$\Rightarrow \: \: \: K_{2top}+U_{2top}=K_{2bottom}+U_{2bottom}$

$\Rightarrow \: \: \: 0+m_{2}gh_{2}=\frac{m_{2}v_{2i}^{2}}{2}+0$

$\Rightarrow \: \: \:v_{2i}=\sqrt{2gh_{2}}$

$\:$

from geometry,

$\:$

$\cos \theta =\frac{L-h_{2}}{L}$

$\Rightarrow \: \: \: L \cos \theta =L-h_{2}$

$\Rightarrow \: \: \: h_{2}=L(1- \cos \theta)$

$\:$

thus,

$\:$

$v_{2i}=\sqrt{2gL(1- \cos \theta)} \: \: \: \: \: \: \: \: \: \: \: \: (3)$

$\:$

inserting eq. (3) into (1), we obtain

$\:$

$v_{1After}=\frac{2m_{2}\sqrt{2gL(1- \cos \theta)}}{m_{1}+m_{2}}$

$\Rightarrow \: \: \: v_{1After}=\frac{2(4.0 \, kg)\sqrt{2(9.8m/s^{2})(3.0m)(1- \cos (60^{\circ}))}}{2.0kg+4.0kg}$

$\:$

or

$\:$

${\color{Blue} v_{1After}=7.23 \, m/s}$

$\:$

inserting eq. (3) into (2), we obtain

$\:$

$v_{2After}=\frac{(m_{2}-m_{1})\sqrt{2gL(1- \cos \theta)}}{m_{1}+m_{2}}$

$v_{2After}=\frac{(4.0kg-2.0kg)\sqrt{2(9.8m/s^{2})(3.0m)(1- \cos (60^{\circ}))}}{2.0kg+4.0kg}$

$\:$

or

$\:$

${\color{Blue} v_{2After}=1.81 \, m/s}$

$\:$

Part (b) - Solution

$\:$

Use again Principle of conservation of mechanical energy to find the maximum height of each ball after the first collision,

$\:$

$E_{bottom}=E_{top}$

$\Rightarrow \: \: \: K_{bottom}+U_{bottom}=K_{top}+U_{top}$

$\Rightarrow \: \: \: \frac{mv_{bot}^{2}}{2}+0=0+mgh_{max}$

$\Rightarrow \: \: \: h_{max}=\frac{v_{bot}^{2}}{2g}$

$\:$

For ball 1, we have

$\:$

$h_{1max}=\frac{v_{1After}^{2}}{2g}$

$\Rightarrow \: \: \: h_{1max}=\frac{(7.23m/s)^{2}}{2(9.8m/s^{2})}$

$\:$

or

$\:$

${\color{Blue} h_{1max}=2.67 \, m}$

$\:$

For ball 2, we have

$\:$

$h_{2max}=\frac{v_{2After}^{2}}{2g}$

$\Rightarrow \: \: \: h_{2max}=\frac{(1.81m/s)^{2}}{2(9.8m/s^{2})}$

$\:$

or

$\:$

${\color{Blue} h_{2max}=0.167 \, m}$