91Ó°ÊÓ

Mass of the first block,$m_1=5\mathrm{kg}$

Initial speed of the first block,$v_{1i}=3\mathrm{m}/\mathrm{s}$

Mass of the first block,$m_2=10\mathrm{kg}$

Initial speed of the first block,$v_{2i}=2\mathrm{m}/\mathrm{s}$

Final speed of second block,$v_{2f}=2.5\mathrm{m}/\mathrm{s}$

For part c, final speed of second block is $v_{2f}=2.5\mathrm{m}/\mathrm{s}$

Usetheprinciple of conservation of momentum and change in initial and final kinetic energy of collision to findtherequired values.According tothe conservation of momentum, momentum of a system is constant if no external forces are acting on the system.

Formulae are as follow:

$$\begin{gathered} P_i=P_f \\ P=mv \\ K=\frac{1}{2}m{v}^2 \\ ∆K=K_f-K_i \end{gathered}$$

where, m is mass, v is velocity, P is linear momentum and K is kinetic energy.

Applying principle of conservation of momentum,

Total momentum$\stackrel{→}{P_i}$before collision = Total momentum after collision$\stackrel{→}{P_f}$

For the given situation,

Total initial momentum = Initial momentum of first block+ Initial momentum of second block.

$$\begin{gathered} \stackrel{→}{P_i}=\stackrel{→}{P_{1i}}+\stackrel{→}{P_{2i}} \\ \stackrel{→}{P_i}=m_1{v}_{1i}+m_2{v}_{2i}................\left(1\right) \end{gathered}$$

Total final momentum = final momentum of first block + final momentum of second block.

$\stackrel{→}{P_f}=m_1{v}_{1f}+m_2{v}_{2f}................\left(2\right)$

Equating equation (1) and (2)

role="math" localid="1661492464511" $$\begin{gathered} m_1{v}_{1i}+m_2{v}_{2i}=m_1{v}_{1f}+m_2{v}_{2f} \\ {m}_1{v}_{1f}=m_1{v}_{1i}+m_2{v}_{2i}-m_2{v}_{2f} \\ {v}_{1f}=\frac{m_1{v}_{1i}+m_2{v}_{2i}-m_2{v}_{2f}}{m_1}............\left(3\right) \\ {v}_{1f}=\frac{5×3+10×2-10×2.5}{5} \\ {v}_{1f}=2.0\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the final speed of first block is 2.0 m/s .

To find change in K.E. due to collision,

Initial K.E. = Initial K.E. of first block + Initial K.E. of second block

$$\begin{gathered} K_i=K_{1i}+K_{2i} \\ {K}_i=\frac{1}{2}{m}_1{v}_{1i}^2+\frac{1}{2}{m}_2{v}_{2i}^2 \\ {K}_i=\frac{1}{2}\left[5×3^2+10×2^2\right] \\ {K}_i=42.5\mathrm{J}..........\left(4\right) \end{gathered}$$

Final K.E. = Final K.E. of first block + Final K.E. of second block

role="math" localid="1661492635653" $$\begin{gathered} K_f=K_{1f}+K_{2f} \\ {K}_f=\frac{1}{2}{m}_1{v}_{1f}^2+\frac{1}{2}{m}_2{v}_{2f}^2 \\ {K}_f=\frac{1}{2}\left[5×2^2+10×{\left(2.5\right)}^2\right] \\ {K}_f=41.25\mathrm{J}..........\left(5\right) \end{gathered}$$

Total change in K.E.,role="math" localid="1661492722767" $∆K=K_f-K_i$

Using equation (4) and (5),

$$\begin{gathered} ∆K=41.25-42.5 \\ ∆K=-1.25\mathrm{J} \\ ≈-1.3\mathrm{J} \end{gathered}$$

Hence, the change in K.E. of the system due to collision is $-1.3\mathrm{J}$

Whenthe final speed ofthe second block is$v_{2f}=4\mathrm{m}/\mathrm{s}$

Using equation (3),

$$\begin{gathered} v_{1f}=\frac{m_1{v}_{1i}+m_2{v}_{2i}-m_2{v}_{2f}}{m_1} \\ {v}_{1f}=\frac{5×3+10×2-10×4}{5} \\ {v}_{1f}=-1.0\mathrm{m}/\mathrm{s} \end{gathered}$$

To find change in K.E. due to collision,

Initial K.E. = Initial K.E. of first block + Initial K.E. of second block

As only final speed ofthesecond block is changed, initial K.E. will remainthe same.

$K_i=42.5\mathrm{J}...........\left(6\right)$

Final K.E. = Final K.E. of first block + Final K.E. of second block

$$\begin{gathered} K_f=K_{1f}+K_{2f} \\ {K}_f=\frac{1}{2}{m}_1{v}_{1f}^2+\frac{1}{2}{m}_2{v}_{2f}^2 \\ {K}_f=\frac{1}{2}\left[5×{\left(-1\right)}^2+10×{\left(4\right)}^2\right] \\ {K}_f=82.5\mathrm{J}..........\left(7\right) \end{gathered}$$

Total change in K.E,$∆K=K_f-K_i$

Using equation (4) and (5),

$$\begin{gathered} ∆K=82.5-42.5 \\ ∆K=40\mathrm{J} \end{gathered}$$

Hence, the change in K.E. of the system due to collision is 40 J .

The additional kinetic energy in part c is possible from some other source.

Therefore, by using the principle of conservation of momentum and change in kinetic energy of the collision, the required answers can be found.