A 0.25 kg puck is initially stationary on an ice surface with negligible friction. At time t = 0, a horizontal force begins to move the puck. The force is given by `vecF=(12.0-3.00t^2)hati`, with `vecF` in newtons and t in seconds, and it acts until its magnitude is zero. (a) What is the magnitude of the impulse on the puck from the force between t= 0.750 s and t = 1.25 s? (b) What is the change in momentum of the puck between t=0 and the instant at which F=0?

Answer

To solve the given problem, we will break it down into two parts: (a) calculating the magnitude of the impulse on the puck from the force between t = 0.750 s and t = 1.250 s, and (b) determining the change in momentum of the puck between t = 0 and the instant at which the force becomes zero.

### Part (a): Magnitude of Impulse

1. **Identify the Force Function**:
The force acting on the puck is given by:
\[ \vec{F} = (12.0 - 3.00t^2) \hat{i} \text{ (in newtons)} \]

2. **Impulse Calculation**:
Impulse (\(J\)) is defined as the integral of force over time:
\[ J = \int_{t_1}^{t_2} \vec{F} \, dt \]
Here, \(t_1 = 0.750\) s and \(t_2 = 1.250\) s.

3. **Set Up the Integral**:
Substitute the force function into the integral:
\[ J = \int_{0.750}^{1.250} (12.0 - 3.00t^2) \, dt \]

4. **Calculate the Integral**:
- First, integrate \(12.0\):
\[ \int 12.0 \, dt = 12.0t \]
- Then, integrate \(-3.00t^2\):
\[ \int -3.00t^2 \, dt = -t^3 \]
- Thus, the integral becomes:
\[ J = \left[ 12.0t - t^3 \right]_{0.750}^{1.250} \]

5. **Evaluate the Integral**:
- Calculate at \(t = 1.250\):
\[ 12.0(1.250) - (1.250)^3 = 15.0 - 1.953125 = 13.046875 \]
- Calculate at \(t = 0.750\):
\[ 12.0(0.750) - (0.750)^3 = 9.0 - 0.421875 = 8.578125 \]
- Now, subtract:
\[ J = 13.046875 - 8.578125 = 4.46875 \text{ Ns} \]

6. **Final Result for Part (a)**:
The magnitude of the impulse is approximately:
\[ J \approx 4.47 \text{ Ns} \]

### Part (b): Change in Momentum

1. **Determine When the Force Becomes Zero**:
Set the force function to zero:
\[ 12.0 - 3.00t^2 = 0 \implies 3.00t^2 = 12.0 \implies t^2 = 4 \implies t = 2.0 \text{ s} \]

2. **Calculate the Change in Momentum**:
The change in momentum (\(\Delta p\)) is equal to the impulse over the entire time from \(t = 0\) to \(t = 2.0\):
\[ \Delta p = \int_{0}^{2.0} \vec{F} \, dt = \int_{0}^{2.0} (12.0 - 3.00t^2) \, dt \]

3. **Set Up the Integral**:
\[ \Delta p = \left[ 12.0t - t^3 \right]_{0}^{2.0} \]

4. **Evaluate the Integral**:
- Calculate at \(t = 2.0\):
\[ 12.0(2.0) - (2.0)^3 = 24.0 - 8.0 = 16.0 \]
- Calculate at \(t = 0\):
\[ 12.0(0) - (0)^3 = 0 \]
- Now, subtract:
\[ \Delta p = 16.0 - 0 = 16.0 \text{ Ns} \]

5. **Final Result for Part (b)**:
The change in momentum of the puck is:
\[ \Delta p = 16.0 \text{ Ns} \]

### Summary of Results
- (a) The magnitude of the impulse on the puck between \(t = 0.750\) s and \(t = 1.250\) s is approximately \(4.47\) Ns.
- (b) The change in momentum of the puck between \(t = 0\) and the instant at which \(F = 0\) is \(16.0\) Ns.