Question

A 5.0 kg block with a speed of 3.0m/scollides with a 10 kg block that has a speed of 2.0 kgin the same direction. After the collision, the10kgblock travels in the original direction with a speed of 2.5 m/s .(a) What is the velocity of the 5.0 kg block immediately after the collision? (b) By how much does the total kinetic energy of the system of two blocks change because of the collision? (c) Suppose, instead, that the 10kg block ends up with a speed of4.0m/s. What then is the change in the total kinetic energy? (d) Account for the result you obtained in (c).

Short answer

1. The final speed of first block is 2.0 m/s .
2. The change in K.E. of the system due to collision is$-1.3\mathrm{J}$
3. The change in K.E. of the system due to collision is 40 J .
4. The additional kinetic energy in part c is possible from some other source.

Step-by-step solution

Step 1: Given Data

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 $v_{2f}=2.5\mathrm{m}/\mathrm{s}$

Determining the concept

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.

(a) Determining the final speed of first block,  vf1

Applying principle of conservation of momentum,

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

For the given situation,

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

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

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

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

Equating equation (1) and (2)

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

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

(b) Determining the change in the total K.E., ∆K

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

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

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

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

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

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

Using equation (4) and (5),

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

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

(c) Determining when the final speed of the second block is 4 m/s

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

Using equation (3),

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

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

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

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 .

(d) Determining the explanation for part c

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.