A `3.0kg` block slides on a frictionless horizontal surface, first moving to the left at `50m//s`. It collides with a spring as it moves left, compresses the spring and is brought to rest momentarily. The body continues to be accelerated to the right by the force of compressed spring. Finally, the body moves to the right at `40m//s`. The block remains in contact with the spring for `0.020s`. What were the magnitude and direction of the impulse of the spring on the block? What was the spring's average force on the block?

Explanation

To solve the problem, we need to find two things: the magnitude and direction of the impulse of the spring on the block, and the average force exerted by the spring on the block.

### Step 1: Determine the Change in Velocity
The block initially moves to the left at a velocity of \( v_i = -50 \, \text{m/s} \) (negative because it is to the left). After compressing the spring, it comes to rest momentarily, and then it moves to the right at a velocity of \( v_f = 40 \, \text{m/s} \) (positive because it is to the right).

### Step 2: Calculate the Change in Velocity
The change in velocity (\( \Delta v \)) can be calculated as:
\[ \Delta v = v_f - v_i = 40 \, \text{m/s} - (-50 \, \text{m/s}) = 40 \, \text{m/s} + 50 \, \text{m/s} = 90 \, \text{m/s} \]

### Step 3: Calculate the Impulse
Impulse (\( J \)) is defined as the product of mass (\( m \)) and change in velocity (\( \Delta v \)):
\[ J = m \cdot \Delta v \]
Given that the mass of the block is \( m = 3.0 \, \text{kg} \):
\[ J = 3.0 \, \text{kg} \cdot 90 \, \text{m/s} = 270 \, \text{kg m/s} \]
The direction of the impulse is to the right (positive direction).

### Step 4: Calculate the Average Force
Impulse can also be expressed in terms of average force (\( F_{\text{avg}} \)) and the time duration (\( \Delta t \)) during which the force acts:
\[ J = F_{\text{avg}} \cdot \Delta t \]
We know that \( \Delta t = 0.020 \, \text{s} \). Rearranging the formula to find the average force gives us:
\[ F_{\text{avg}} = \frac{J}{\Delta t} = \frac{270 \, \text{kg m/s}}{0.020 \, \text{s}} = 13500 \, \text{N} \]
Thus, the average force exerted by the spring on the block is \( 13500 \, \text{N} \) to the right.

### Final Answers
1. **Magnitude and Direction of Impulse**: \( 270 \, \text{N s} \) to the right.
2. **Average Force of the Spring on the Block**: \( 13500 \, \text{N} \) to the right.