Question

Particle 1 of mass 200g and speed$\mathbf{3}\mathbf{.}\mathbf{00}\frac{\mathbf{m}}{\mathbf{s}}$undergoes a one dimensional collision with stationary particle 2 of mass 400g.What is the magnitude of the impulse on particle 1 if the collision is (a) elastic and (b) completely inelastic?

Short answer

1. The magnitude of the impulse on particle 1 if the collision is elastic is$0.8\frac{\mathrm{kg}}{\mathrm{s}}$.
2. The magnitude of the impulse on particle 1 if the collision is completely inelastic is $0.4\frac{\mathrm{kg}·\mathrm{m}}{\mathrm{s}}$.

Step-by-step solution

Step 1: Given

1. Mass of the first particle,$m_1=200\mathrm{gm}=0.2\mathrm{kg}$
2. Initial speed of the first particle,$v_{1i}=3.00\frac{\mathrm{m}}{\mathrm{s}}$
3. Mass of the second particle,$m_2=400\mathrm{gm}=0.4\mathrm{kg}$
4. Initial speed of the second particle,$v_{2i}=0.00\frac{\mathrm{m}}{\mathrm{s}}$

Determining the concept

When collision is completely elastic, use the conservation of momentum and conservation of energy to find the impulse on particle 1. If the collision is completely inelastic, then conservation of momentum holds good.According tothe conservation of momentum, momentum of a system is constant if no external forces are acting on the system.

Formulae are as follow:

$$\begin{gathered} m_1{v}_{1i}+m_2{v}_{2i}=m_1{v}_{1f}+m_2{v}_{2f} \\ ∆p=\left|p_f-p_i\right|=\left|m_1{v}_{1f}-m_1{v}_{1i}\right| \end{gathered}$$

Here, $m_1$, and $m_2$ are masses, vis velocity and p is momentum.

(a) Determine the magnitude of the impulse on particle 1 if the collision is elastic

For conservation of momentum,

$$\begin{gathered} m_1{v}_{1i}+m_2{v}_{2i}=m_1{v}_{1f}+m_2{v}_{2f} \\ \left(0.2\right)\left(3.00\frac{\mathrm{m}}{\mathrm{s}}\right)+\left(0.4\mathrm{kg}\right)\left(0.00\frac{\mathrm{m}}{\mathrm{s}}\right)=\left(0.2\mathrm{kg}\right)v_{1f}+\left(0.4\mathrm{kg}\right)v_{2f} \\ \left(0.6\frac{\mathrm{kg}·\mathrm{m}}{\mathrm{s}}\right)=\left(0.2\mathrm{kg}\right)v_{1f}+\left(0.4\mathrm{kg}\right)v_{2f} \\ {v}_{1f}+2v_{2f}=3.00\frac{m}{s} \end{gathered}$$

$v_{1f}=3.00\frac{\mathrm{m}}{\mathrm{s}}-2v_{2f}$ …… (1)

Now, for conservation of kinetic energy write the equation as:

$\frac{1}{2}{m}_1{v}_{1i}^2+\frac{1}{2}{m}_2{v}_{1f}^2=\frac{1}{2}{m}_1{v}_{1f}^2+\frac{1}{2}{m}_1{v}_{2f}^2$

Substitute the values and solve as:

$$\begin{gathered} \left(0.2kg\right){\left(3.00\frac{\mathrm{m}}{\mathrm{s}}\right)}^2+\left(0.4kg\right){\left(0.00\frac{\mathrm{m}}{\mathrm{s}}\right)}^2=\left(0.2\mathrm{kg}\right)v_{1f}^2+\left(0.4\mathrm{kg}\right)v_{2f}^2 \\ 1.8kg\frac{{\mathrm{m}}^2}{{\mathrm{s}}^2}=\left(0.2\mathrm{kg}\right)v_{1f}^2+\left(0.4\mathrm{kg}\right)v_{2f}^2 \end{gathered}$$

$v_{1f}^2+2v_{2f}^2=9.00$ …… (2)

Use equation 1 into equation 2 solve further as:

$$\begin{gathered} {\left(3.00\frac{\mathrm{m}}{\mathrm{s}}-2v_{2f}\right)}^2+2v_{2f}^2=9.00 \\ 9.00-12.00v_{2f}+4v_{2f}^2=9.00 \\ 6v_{2f}^2=12.00v_{2f} \\ {v}_{2f}=2.00\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Using this value into equation 1 solve as:

$$\begin{gathered} v_{1f}=3.00\frac{\mathrm{m}}{\mathrm{s}}-2\left(2.00\frac{\mathrm{m}}{\mathrm{s}}\right) \\ {v}_{1f}=-1.00\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Therefore,

$$\begin{gathered} ∆p=\left|p_f-p_i\right|=\left|m_1{v}_{1i}\right| \\ ∆p=\left|\left(0.2\mathrm{kg}\right)\left(-1.00\frac{\mathrm{m}}{\mathrm{s}}\right)-\left(0.2\mathrm{kg}\right)\left(3.00\frac{\mathrm{m}}{\mathrm{s}}\right)\right| \\ ∆p=0.8\frac{\mathrm{kg}·\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Therefore, the change in ball’s momentum is$8.0\frac{\mathrm{kg}·\mathrm{m}}{\mathrm{s}}$

(b) Determine the magnitude of the impulse on particle 1 if the collision is completely inelastic

For a completely inelastic collision:

$m_1{v}_{1i}+m_2{v}_{2i}=\left(m_1+m_2\right)v_{1f}$

Substitute the values and solve as:

$$\begin{gathered} \left(0.2\mathrm{kg}\right)\left(3.00\frac{\mathrm{m}}{\mathrm{s}}\right)+\left(0.4\mathrm{kg}\right)\left(0.00\frac{\mathrm{m}}{\mathrm{s}}\right)=\left[\left(0.2\mathrm{kg}\right)+\left(0.4\mathrm{kg}\right)\right]V \\ {v}_{1f}=\frac{\left(0.2\mathrm{kg}\right)\left(3.00{\displaystyle \frac{\mathrm{m}}{\mathrm{s}}}\right)}{\left(2.00\mathrm{kg}\right)+\left(0.4\mathrm{kg}\right)} \\ {v}_{1f}=1.0\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Therefore, solve for the change in momentum as:

$$\begin{gathered} ∆p=\left|p_f-p_i\right| \\ =\left|m_1{v}_{1f}-m_1{v}_{1i}\right| \end{gathered}$$

Substitute the values and solve as:

$$\begin{gathered} ∆p=\left|\left(0.2\mathrm{kg}\right)\left(1.00\frac{\mathrm{m}}{\mathrm{s}}\right)-\left(0.2\mathrm{kg}\right)\left(3.00\frac{\mathrm{m}}{\mathrm{s}}\right)\right| \\ ∆p=0.4\frac{\mathrm{kg}·\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Hence, the magnitude of the impulse on particle 1 if the collision is completely inelastic, is$0.4\mathrm{kg}\frac{\mathrm{m}}{\mathrm{s}}$.

Therefore, the magnitude of the impulse on particle 1 can be found if the collision is elastic using the conservation of momentum and conservation of kinetic energy. Also, the magnitude of the impulse on particle 1 can be found if the collision is completely inelastic using the conservation of momentum only because the momentum is conserved in an inelastic collision.