Solve the physics questions related to mechanics, momentum, and collisions, including questions about elastic vs inelastic collisions, railway truck collisions, and truck/car colli... Solve the physics questions related to mechanics, momentum, and collisions, including questions about elastic vs inelastic collisions, railway truck collisions, and truck/car collisions. Read more

Steps to Solve

1. Find mass of clay from previous steps

Since this question builds upon previous ones, we need the mass of the clay. Let's assume from a previous part (not shown here) that the mass of the clay is $0.3 \text{ kg}$

2. Calculate the initial momentum of the clay

The clay is initially at rest. Momentum $p$ is given by $p = mv$, where $m$ is mass and $v$ is velocity. Since the clay is stationary, its initial velocity is $0 \text{ m/s}$. Therefore, the initial momentum of the clay is $p = (0.3 \text{ kg}) \times (0 \text{ m/s}) = 0 \text{ kg m/s}$.

3. Calculate the momentum of the air-rifle pellet before hitting the clay.

We need the mass and velocity of the pellet. Let's assume from a previous part (not shown here) that the mass of the pellet is $0.002 \text{ kg}$ (2 grams) and the velocity is $725 \text{ m/s}$. Then the momentum of the pellet before impact is $p = (0.002\text{ kg}) \times (725 \text{ m/s}) = 1.45 \text{ kg m/s}$.

4. State the principle of conservation of linear momentum.

The principle of conservation of linear momentum states that the total momentum of an isolated system remains constant if no external forces act on it. In other words, in a closed system, the total momentum before a collision is equal to the total momentum after the collision.

5. Show that the speed of the clay block, with the pellet inside, is $4.83 \text{ m/s}$ after the collision.

Let $v$ be the speed of the clay block with the pellet inside after the collision. Using conservation of momentum:

$$ \text{Momentum before} = \text{Momentum after} $$

Momentum of pellet before + Momentum of clay before = Momentum of (pellet + clay) after

$$ (0.002 \text{ kg} \times 725 \text{ m/s}) + (0.3 \text{ kg} \times 0 \text{ m/s}) = (0.002 \text{ kg} + 0.3 \text{ kg}) \times v $$

$$ 1.45 \text{ kg m/s} = (0.302 \text{ kg}) \times v $$

$$ v = \frac{1.45 \text{ kg m/s}}{0.302 \text{ kg}} \approx 4.80 \text{ m/s} $$

Rounding to two decimal places as given in the problem makes the speed $4.80 \text{ m/s}$.

Note: A discrepancy exists between the answer $4.83 \text{ m/s}$ provided in the question and the value $4.80 \text{ m/s}$ calculated here using the conservation of momentum and assuming that the provided values are correct ($m_{pellet}=0.002 \text{ kg}$, $v_{pellet}=725 \text{ m/s}$, and $m_{clay}=0.3 \text{ kg}$). It is possible that the values are rounded differently in the original problem leading to this difference, or the value reported in the problem is slightly off.

6. Determine whether the collision is elastic or inelastic.

To determine if the collision is elastic or inelastic, we need to compare the kinetic energy before and after the collision.

Kinetic energy before: $$KE_{before} = \frac{1}{2} m_{pellet} v_{pellet}^2 + \frac{1}{2} m_{clay} v_{clay}^2$$ Using the values from the previous steps $$KE_{before} = \frac{1}{2} (0.002 \text{ kg}) (725 \text{ m/s})^2 + \frac{1}{2} (0.3 \text{ kg}) (0 \text{ m/s})^2 = 525.625 \text{ J}$$

Kinetic energy after: $$KE_{after} = \frac{1}{2} (m_{pellet} + m_{clay}) v^2$$ $$KE_{after} = \frac{1}{2} (0.002 \text{ kg} + 0.3 \text{ kg}) (4.80 \text{ m/s})^2$$

$$KE_{after} = \frac{1}{2} (0.302 \text{ kg}) (4.80 \text{ m/s})^2 \approx 3.47 \text{ J}$$

Since $KE_{before} \neq KE_{after}$ (specifically $KE_{before} > KE_{after}$), the collision is inelastic because kinetic energy is not conserved.