Chapter 8, Problem 57A

A 2-kg blob of putty moving at 3 m/s slams into a 2-kg blob of putty at rest.

a. Calculate the speed of the two stuck-together blobs of putty immediately after colliding.

b. Calculate the speed of the two blobs if the one at rest was 4 kg.

To determine

To Calculate: The speed of two stuck-together blobs.

Answer to Problem 57A

  $1.5\text{m/s}$

Explanation of Solution

Given:

The mass of moving blob of putty is 2kg and its initial velocity is $3\text{m/s}$ and the mass of blob of putty at rest is 2kg.

Formula used:

According to the law of conservation of momentum,

Initial momentum = final momentum

The momentum is given as,

P=mv

As per law of conservation of momentum is:

  $m_1{v}_{1b}+m_2{v}_{2b}=m_1{v}_{1a}+m_2{v}_{2a}$

Where, $m_1\text{and}{m}_2$ are masses of blob and $v_{1b}\text{and}{v}_{2b}$ are velocities of the particles before collision and $v_{1a}\text{and}{v}_{2a}$ are the velocities after collision.

Calculation:

According to the law of conservation of momentum, the system momentum remains conserved if no external force is acting on the system. After the collision, the velocities become equal. So, the formula becomes,

  $m_1{v}_{1b}+m_2{v}_{2b}=\left(m_1+m_2\right)v_A$

Substitute the values and solve:

  $\begin{array}{l}{m}_1{v}_{1b}+m_2{v}_{2b}=\left(m_1+m_2\right)v_A\\ 2\times 3+0=\left(2+2\right)v_A\\ \Rightarrow v_a=\frac{6}{4}=1.5\text{m/s}\end{array}$

Conclusion:

Thus, the velocity is $1.5\text{m/s}$ .

To determine

To Calculate: The speed of the two blobs.

Answer to Problem 57A

  $1\text{m/s}$

Explanation of Solution

Given information:

The blob which is at rest has the mass 4kg.

Formula used:

As per law of conservation of momentum is:

  $m_1{v}_{1b}+m_2{v}_{2b}=m_1{v}_{1a}+m_2{v}_{2a}$

Where, $m_1\text{and}{m}_2$ are masses of blob and $v_{1b}\text{and}{v}_{2b}$ are velocities of the particles before collision and $v_{1a}\text{and}{v}_{2a}$ are the velocities after collision.

after collision.

Calculation:

According to the law of conservation of momentum, the system momentum remains conserved if no external force is acting on the system. After the collision, the velocities become equal. So, the formula becomes,

  $m_1{v}_{1b}+m_2{v}_{2b}=\left(m_1+m_2\right)v_A$

By substituting the values in the formula, the velocity after the collision is

  $\begin{array}{l}{m}_1{v}_{1b}+m_2{v}_{2b}=\left(m_1+m_2\right)v_A\\ 2\times 3+0=\left(2+4\right)v_A\\ \Rightarrow v_a=\frac{6}{6}=1\text{m/s}\end{array}$

Conclusion:

Thus, the speed is $1\text{m/s}$ .