Chapter 7, Problem 39P

(a)

To determine

Final velocities of both mass after collision.

Answer to Problem 39P

Both objects after a head-on collision will stick together and move with the same velocity that is $v_1=v_2=1.7\text{m/s}$

Explanation of Solution

Given:

Object mass is 15 kg. It moves in +x direction at 5.5 m/s. it collides with a mass of 10 kg that is travelling in -x direction at 4 m/s.

After the collision, objects stick together that is they move with equal velocity as a single unit $v_1=v_2$

Formula Used:

According to the conservation of momentum,

  $m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$

Calculation:

Substituting the values in the formula for the conservation of momentum,

  $\begin{array}{l}{m}_1{u}_1+m_2{u}_2=(m_1+m_2)v_1\\ v_1=\frac{m_1{u}_1+m_2{u}_2}{(m_1+m_2)}\\ v_1=\frac{(15\times 5.5)+[10\times \left(-4\right)]}{(15+10)}=1.7\text{m/s}\end{array}$

Conclusion:

Thus, both objects after a head-on collision will stick together and move with the same velocity that is $v_1=v_2=1.7\text{m/s}$

(b)

To determine

Final velocities of both mass after collision.

Answer to Problem 39P

After the collision, object with mass $m_1$ will move in −x direction with velocity 2.1 m/s and object with mass $m_2$ will move in +x direction with velocity 7.4 m/s

Explanation of Solution

Given:

Object mass is 15 kg. It moves in +x direction at 5.5 m/s. it collides with a mass of 10 kg that is travelling in -x direction at 4 m/s.

The collision is elastic which means,

  $\begin{array}{l}{u}_1-u_2=-(v_1-v_2)\\ v_2=u_1-u_2+v_1\end{array}$

Formula Used:

According to the conservation of momentum,

  $m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$

Calculation:

After inserting the values in the formula for the conservation of momentum,

  $\begin{array}{l}{m}_1{u}_1+m_2{u}_2=m_1{v}_1+m_2(u_1-u_2+v_1)\\ v_1=\frac{(m_1-m_2)u_1+2m_2{u}_2}{(m_1+m_2)}\\ v_1=\frac{(15-10)\times 5.5+[2\times 10\times \left(-4\right)]}{(15+10)}=-2.1\text{m/s}\\ v_2=(u_1-u_2+v_1)\\ v_2=5.5+4-2.1=7.4\text{m/s}\end{array}$

Conclusion:

Thus, after the collision, object with mass $m_1$ will move in −x direction with velocity 2.1 m/s, and object with mass $m_2$ will move in +x direction with velocity 7.4 m/s

(c)

To determine

Final velocities of both mass after collision.

To identify: Whether the result obtained is reasonable.

Answer to Problem 39P

Both objects after a head-on collision, 15 kg object is at rest that is $v_1=0\text{m/s}$ and 10 kg object will move in +x direction with $v_2=4.3\text{m/s}$ .

Yes, it is reasonable.

Explanation of Solution

Given:

Object mass is 15 kg. It moves in +x direction at 5.5 m/s. it collides with a mass of 10 kg that is travelling in -x direction at 4 m/s.

After the collision, 15 kg object is at rest.

Formula Used:

According to the conservation of momentum,

  $m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$

According to the conservation of kinetic energy,

  $\Delta KE=\frac{1}{2}(m_1{v}^2{}_1+m_2{v}^2{}_2)-\frac{1}{2}(m_1{u}^2{}_1+m_2{u}^2{}_2)$

Calculation:

After inserting the values in the formula for the conservation of momentum, we get

  $\begin{array}{l}{m}_1{u}_1+m_2{u}_2=m_2{v}_2\\ v_2=\frac{m_1{u}_1+m_2{u}_2}{m_2}\\ v_2=\frac{(15\times 5.5)+[10\times \left(-4\right)]}{10}=4.3m/s\end{array}$

Now, calculating change in kinetic energy,

  $\begin{array}{l}\Delta KE=\frac{1}{2}{m}_2{v}^2{}_2-\frac{1}{2}(m_1{u}^2{}_1+m_2{u}^2{}_2)\\ \Delta KE=\left[\frac{1}{2}(10){(4.3)}^2-\frac{1}{2}(15){(5.5)}^2-\frac{1}{2}(10){(-4)}^2\right]\\ \Delta KE=-220\text{J}\end{array}$

Here, system has lost kinetic energy.

Now, after checking the direction of the system after collision, $10 kg$ object will move in +x direction and $15 kg$ object will stop by losing kinetic energy. This suggests that the result is reasonable.

Conclusion:

Both objects after a head-on collision, 15 kg object is at rest that is 0 m/s and 10 kg object will move in +x direction with $v_2=4.3\text{m/s}$

The result is reasonable.

(d)

To determine

Final velocities of both mass after collision.

To identify: Whether the result obtained is reasonable.

Answer to Problem 39P

Both objects after a head-on collision, 10 kg object is at rest that is $v_2=0\text{m/s}$ and 15 kg object will move in +x direction with $v_1=2.8\text{m/s}$ . The result is not reasonable.

Explanation of Solution

Given:

Object mass is 15 kg. It moves in +x direction at 5.5 m/s. it collides with a mass of 10 kg that is travelling in -x direction at 4 m/s.

After the collision, 10 kg object is at rest that is $v_2=0\text{m/s}$ .

Formula Used:

According to the conservation of momentum,

  $m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$

Calculation:

After inserting the values in the formula for the conservation of momentum,

  $\begin{array}{l}{m}_1{u}_1+m_2{u}_2=m_1{v}_1\\ v_1=\frac{m_1{u}_1+m_2{u}_2}{m_1}\\ v_1=\frac{(15\times 5.5)+[10\times \left(-4\right)]}{15}=2.8\text{m/s}\end{array}$

Now, after checking the direction of the system, after collision, 15 kg object will move in +x direction and 10 kg object will be stopped. But, 15 kg object cannot pass the other object moving in the +x direction. If 10 kg object stops, then 15 kg object must move in −x direction. Hence, this result is not reasonable.

Conclusion:

Both objects after a head-on collision, 10 kg object is at rest that is $v_2=0\text{m/s}$ and 15 kg object will move in +x direction with $v_1=2.8\text{m/s}$ . The result is not reasonable.

(e)

To determine

Final velocities of both mass after collision.

To identify: Whether the result obtained is reasonable.

Answer to Problem 39P

Both objects after a head-on collision, 10 kg object is moving in +x direction with velocity $v_2=10.3\text{m/s}$ and 15 kg object will move in -x direction with $v_1=4\text{m/s}$ . The system is not reasonable.

Explanation of Solution

Given:

Object mass is 15 kg. It moves in +x direction at 5.5 m/s. it collides with a mass of 10 kg that is travelling in -x direction at 4 m/s.

After the collision, 15 kg object is moving in negative direction with velocity $v_1=4\text{m/s}$

Formula Used:

According to the conservation of momentum,

  $m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$

Calculation:

After substituting the values in the formula for the conservation of momentum,

  $\begin{array}{l}{m}_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2\\ v_2=\frac{m_1{u}_1+m_2{u}_2-m_1{v}_1}{m_2}\\ v_2=\frac{(15\times 5.5)+[10\times \left(-4\right)]-[15\times \left(-4\right)]}{10}=10.3\text{m/s}\end{array}$

Now, after checking the direction of the system, directions of the system is accurate. But, considering the case of perfectly elastic collision, velocity of both objects is larger in this case. This suggests that both particles have gained kinetic energy which is not possible without any external source or greater initial velocities. Hence, result is not reasonable.

Conclusion:

Both objects after a head-on collision, 10 kg object is moving in +x direction with velocity $v_2=10.3\text{m/s}$ and 15 kg object will move in -x direction with $v_1=4\text{m/s}$ . The result is not reasonable.