AP Physics C Mechanics - 3.1 Translational Kinetic Energy- Exam Style questions- FRQs

(a)
The net impulse is defined as the integral of the net force over the given time duration:
\( J = \int F \, dt = \int m a(t) \, dt \)
\( J = m \int_{0}^{6.0} (9.0 – 0.25t^2) \, dt \)
\( J = 0.50 \left[ 9.0t – \frac{0.25}{3}t^3 \right]_{0}^{6.0} \)
\( J = 0.50 \left[ 9.0(6.0) – \frac{1}{12}(6.0)^3 \right] \)
\( J = 0.50 [ 54 – 18 ] = 0.50 (36) \)
\( \boxed{J = 18\,\text{N}\cdot\text{s}} \)

(b)
By the impulse-momentum theorem, the net impulse equals the change in linear momentum:
\( J = \Delta p = m(v_f – v_i) \)
Since the toy rocket starts from rest, the initial velocity is zero:
\( 18\,\text{N}\cdot\text{s} = 0.50\,\text{kg} \times (v_f – 0) \)
\( v_f = \frac{18}{0.50} \)
\( \boxed{v_f = 36\,\text{m/s}} \)

(c) i.
The kinetic energy at the end of the powered phase is evaluated using the formula:
\( K = \frac{1}{2} m v^2 \)
Substituting the velocity calculated in part (b):
\( K = \frac{1}{2} (0.50\,\text{kg}) (36\,\text{m/s})^2 \)
\( K = 0.25 \times 1296 \)
\( \boxed{K = 324\,\text{J}} \)

ii.
To find the vertical displacement during the first six seconds, we integrate the velocity function:
\( v(t) = \int a(t) \, dt = \int (9.0 – 0.25t^2) \, dt = 9.0t – \frac{1}{12}t^3 \)
\( \Delta y = \int_{0}^{6.0} v(t) \, dt = \int_{0}^{6.0} \left( 9.0t – \frac{1}{12}t^3 \right) \, dt \)
\( \Delta y = \left[ \frac{9.0}{2}t^2 – \frac{1}{48}t^4 \right]_{0}^{6.0} \)
\( \Delta y = \left[ 4.5(6.0)^2 – \frac{1}{48}(6.0)^4 \right] = 162 – 27 = 135\,\text{m} \)
Now, calculate the change in gravitational potential energy using \(g = 9.8\,\text{m/s}^2\):
\( \Delta U_g = m g \Delta y = (0.50\,\text{kg})(9.8\,\text{m/s}^2)(135\,\text{m}) \)
\( \boxed{\Delta U_g = 660\,\text{J}} \)

(d)
After the fuel burns out, the rocket acts as a free projectile moving upward against gravity until it stops momentarily at its highest peak. We can apply conservation of mechanical energy or vertical kinematics for this unpowered segment:
\( v_{\text{top}}^2 = v_f^2 + 2 (-g) \Delta y_{\text{free}} \)
\( 0 = (36)^2 – 2(9.8)\Delta y_{\text{free}} \)
\( \Delta y_{\text{free}} = \frac{1296}{19.6} \approx 66\,\text{m} \)
The maximum total height relative to the ground is the sum of the positions from both stages:
\( y_{\text{max}} = \Delta y + \Delta y_{\text{free}} = 135\,\text{m} + 66\,\text{m} \)
\( \boxed{y_{\text{max}} = 201\,\text{m}} \)
(Note: Using \(g = 10\,\text{m/s}^2\) gives \(135 + 64.8 = 199.8\,\text{m}\))

(e)
The graph contains two clear distinct mathematical intervals. For \(0 \le t \le 6.0\,\text{s}\), the curve begins at the origin and scales up as a concave-down cubic curve because acceleration decreases over time, topping out exactly at a maximum of \((6.0\,\text{s}, 36\,\text{m/s})\). After \(t = 6.0\,\text{s}\), it turns instantly into a straight line with a constant negative slope equal to \(-g\), crossing the time axis at \(T_{\text{top}}\) and continuing downward into negative territory until the rocket impacts the ground.