AP Physics 1 FRQ Practice

🎡AP Physics 1

Practice FRQ 1 of 191/19

2. A student launches a small ball from the edge of a level platform while standing on a cart that moves at a constant velocity along a straight, level track, as shown in Figure 1.

Figure 1. Launch of a ball from a cart moving at constant speed: ground frame (a) and cart frame (b).

Figure 2. Vertical velocity in the ground frame, v_y versus time t, for the first 1.0 s after launch.

Figure 3. Horizontal velocity in the ground frame, v_x versus time t, for the first 1.0 s after launch.

Draw and label the ball's velocity components in the ground frame. Consider the ball's motion as measured in the ground frame. At $t=0$ , the ball's velocity relative to the cart has components $v'_{0x} = v_0\cos\theta$ and $v'_{0y} = v_0\sin\theta$ . The cart moves at $v_c=2.0\ \text{m/s}$ to the right relative to the ground. Air resistance is negligible.

On Figure 2, draw a graph of $v_y$ versus $t$ for $0 ≤ t ≤ 1.0\ \text{s}$ .
On Figure 3, draw a graph of $v_x$ versus $t$ for $0 ≤ t ≤ 1.0\ \text{s}$ .

On each graph:
• Clearly label the value at $t=0$ .
• Indicate the slope.
• Use the numerical values given in the problem (including $g$ ) to determine the correct intercepts and slopes.

Starting with the kinematic equations for projectile motion, derive an expression for the horizontal range $R$ (the horizontal distance in the ground frame from $x=0$ to where the ball hits the ground). Express your final answer in terms of $v_0$ , $\theta$ , $v_c$ , $y_0$ , and $g$ , as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information. In the ground frame, the ball is launched from $x=0$ and $y_0=1.20\ \text{m}$ at $t=0$ . The initial speed relative to the cart is $v_0=6.0\ \text{m/s}$ at $\theta=40^\circ$ above horizontal, and the cart moves at $v_c=2.0\ \text{m/s}$ to the right relative to the ground.

Figure 4. Ground-frame trajectory: y versus x for the ball launched from height 1.20 m.

Figure 5. Cart-frame trajectory: y′ versus x′ from launch until the ball hits the ground at y′ = −1.20 m.

Now consider the ball's motion as measured in the cart frame, which is an inertial reference frame because the cart moves at constant velocity. In the cart frame, the ball is launched from $x'=0$ and $y'=0$ at $t=0$ with initial components $v'_{0x}=v_0\cos\theta$ and $v'_{0y}=v_0\sin\theta$ . The vertical acceleration is $-g$ in both frames.

Sketch and label the trajectory of the ball in the cart frame on Figure 5 from launch until it hits the ground (where $y'=-1.20\ \text{m}$ ).

ii.

Sketch and label the trajectory of the ball in the ground frame on Figure 4 over the same time interval. On your sketch, show the initial velocity vector in the ground frame and label its horizontal and vertical components $v_{0x}$ and $v_{0y}$ .

Figure 6. Displacement of the ball in the ground frame over the first 0.50 s.

Indicate whether the magnitude of the displacement of the ball during the interval $0≤ t ≤ 0.50\ \text{s}$ is greater than, less than, or equal to the magnitude of the displacement during the interval $0.50\ \text{s}≤ t ≤ 1.00\ \text{s}$ . Use the ground frame. Consider the time interval from $t=0$ to $t=0.50\ \text{s}$ , as shown in Figure 6. The given values are: $v_0=6.0\ \text{m/s}$ , $\theta=40^\circ$ , $v_c=2.0\ \text{m/s}$ , $g=9.8\ \text{m/s}^2$ , and $\Delta t = 0.50\ \text{s}$ .

first interval displacement > second interval displacement
first interval displacement < second interval displacement
first interval displacement = second interval displacement
Justify how your response is consistent with the velocity-time graphs you drew in part A and the trajectories you sketched in part C.

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