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Let's solve each part of this problem step by step.

### Given Data
- Mass of steel ball \( m = 5.21 \, \text{g} = 0.00521 \, \text{kg} \)
- Initial speed of steel ball \( v = 358 \, \text{m/s} \)
- Mass of steel block \( M = 14.8 \, \text{kg} \)
- The block is initially at rest, so its initial velocity \( V = 0 \).

The collision is perfectly elastic, which means both momentum and kinetic energy are conserved.

### Part (a) - Speed of the Block After the Collision

In a perfectly elastic collision between two objects, the final velocity of each object can be calculated with the following formulae:

For the steel ball (object 1, with mass \( m \) and initial velocity \( v \)):
\[
v' = \frac{(m - M)v + 2MV}{m + M}
\]

For the steel block (object 2, with mass \( M \) and initial velocity \( V \)):
\[
V' = \frac{(M - m)V + 2mv}{m + M}
\]

Since \( V = 0 \), the equation for \( V' \) simplifies to:
\[
V' = \frac{2mv}{m + M}
\]

Substitute the values:
\[
V' = \frac{2 \cdot 0.00521 \, \text{kg} \cdot 358 \, \text{m/s}}{0.00521 \, \text{kg} + 14.8 \, \text{kg}}
\]

Calculate \( V' \):
\[
V' = \frac{3.7276}{14.80521} \approx 0.2517 \, \text{m/s}
\]

So, the speed of the block immediately after the collision is approximately \( 0.2517 \, \text{m/s} \).

### Part (b) - Impulse Exerted on the Block

Impulse \( J \) can be calculated as the change in momentum of the block:
\[
J = \Delta p = M V' - M V
\]

Since \( V = 0 \), this simplifies to:
\[
J = M V'
\]

Using \( V' = 0.2517 \, \text{m/s} \):
\[
J = 14.8 \, \text{kg} \cdot 0.2517 \, \text{m/s} \approx 3.7222 \, \text{kg} \cdot \text{m/s}
\]

Thus, the impulse exerted on the block is approximately \( 3.7222 \, \text{kg} \cdot \text{m/s} \).

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Would you like further details on any of these steps, or have any questions?

Here are some related questions to explore:

1. How would the results change if the collision were inelastic?
2. How would we calculate the speed of the ball after the collision?
3. What is the significance of conservation of kinetic energy in elastic collisions?
4. How can impulse be related to the force exerted during the collision?
5. How would increasing the mass of the block affect the final velocities?

**Tip:** For elastic collisions, remember that both momentum and kinetic energy are conserved, which can simplify calculations significantly.

A steel ball with mass m = 5.21 g is moving horizontally with speed v = 358 m/s when it strikes a block of hardened steel with mass M = 14.8 kg (initially at rest). The ball bounces off the block in a perfectly elastic collision. (a) What is the speed of the block immediately after the collision? (b) What is the impulse exerted on the block?

Learn how to calculate the speed of a block after an elastic collision with a steel ball and determine the impulse exerted on the block. This problem explores conservation of momentum and kinetic energy with detailed formulas and calculations. Suitable for Grades 11-12 physics students.

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