91Ó°ÊÓ

(a)

Given Data:

The velocity of sphere A before the collision is $u_a=1.50\mathrm{m}/\mathrm{s}$

The velocity of sphere B before the collision is $u_b=0.50\mathrm{m}/\mathrm{s}$

The velocity of all the spheres after the collision is $u=0.50\mathrm{m}/\mathrm{s}$

The mass of sphere A is: $m_a=0.020\mathrm{kg}$

The mass of sphere B is: $m_b=0.030\mathrm{kg}$

The mass of sphere C is: $m_c=0.050\mathrm{kg}$

The angle of sphere B with horizontal is $\mathrm{θ}=60°$

The conservation of momentum before and after the collision of all spheres along horizontal and vertical directions is used to find the horizontal and vertical components of the velocity of sphere C.

Apply the conservation of momentum to find the horizontal component of velocity of sphere C.

$m_c{u}_{cx}-m_a{u}_a-m_b{u}_b\cos \mathrm{θ}=\left(m_a+m_b+m_c\right)v$

$u_{cx}$is the horizontal component of velocity of sphere C before the collision.

$$\begin{gathered} \left[\begin{array}{c}\left(0.050\mathrm{kg}\right){\mathrm{u}}_{\mathrm{cx}}-\left(0.020\mathrm{kg}\right)\left(1.50\mathrm{m}/\mathrm{s}\right)\\ -\left(0.030\mathrm{kg}\right)\left(0.50\mathrm{m}/\mathrm{s}\right)\left(\cos 60°\right)\end{array}\right]=\left(0.020\mathrm{kg}+0.030\mathrm{kg}+0.050\mathrm{kg}\right)\left(0.50\mathrm{m}/\mathrm{s}\right) \\ {\mathrm{u}}_{\mathrm{cx}}=1.75\mathrm{m}/\mathrm{s} \end{gathered}$$

Apply the conservation of momentum to find the vertical component of velocity of sphere C.

$m_c{u}_{cy}=m_b{u}_b\sin \mathrm{θ}$

$u_{cy}$is the vertical component of velocity of sphere C before collision.

$$\begin{gathered} \left(0.050kg\right)u_{cy}=\left(0.030\mathrm{kg}\right)\left(0.50\mathrm{m}/\mathrm{s}\right)\left(\sin 60°\right) \\ {u}_{cy}=0.260\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the x and y component of sphere C before collision are 1.75 m/s and 0.260 m/s .

(b)

The change in kinetic energy of the system of spheres is given as:

$∆K=\frac{1}{2}\left(m_a+m_b+m_c\right)v^2-\left[\frac{1}{2}{m}_a{u}_a^2+\frac{1}{2}{m}_b{u}_c^2+\frac{1}{2}\left(u_{cx}^2+u_{cy}^2\right)\right]$

Substitute all the values in the above equation.

role="math" localid="1667977744920" $$\begin{gathered} ∆K=\frac{1}{2}\left(0.020kg+0.030kg+0.050kg\right){\left(0.50m/s\right)}^2-\left[\begin{array}{c}\frac{1}{2}\left(0.020kg\right){\left(1.50m/s\right)}^2\\ +\frac{1}{2}\left(0.030kg\right){\left(0.50m/s\right)}^2\\ +\frac{1}{2}\left(0.050kg\right){\left(1.75m/s\right)}^2+{\left(0.260m/s\right)}^2\end{array}\right] \\ ∆K=-0.092\mathrm{J} \end{gathered}$$

Therefore, the change in kinetic energy of system of spheres is $-0.092\mathrm{J}$.