Vectors and Motion in Two Dimensions Practice

Sample questions

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1. Sample 1difficulty 2/5

   For a fixed launch speed on level ground (no air resistance), the angle that maximizes range is

   • A

     $30^\circ$

   • B

     $60^\circ$

   • C

     $45^\circ$

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   • D

     $90^\circ$

   Why

   $R = v_0^2 \sin(2\theta)/g$ is maximum when $\sin(2\theta) = 1$, i.e. $\theta = 45^\circ$.

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2. Sample 2difficulty 2/5

   A marble rolls off a horizontal table $1.25~\text{m}$ above the floor with a horizontal speed of $4.0~\text{m/s}$. Using $g \approx 10~\text{m/s}^2$ and ignoring air resistance, how far from the table edge does it land?

   • A

     $4.0~\text{m}$

   • B

     $2.0~\text{m}$

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   • C

     $3.0~\text{m}$

   • D

     $1.0~\text{m}$

   Why

   Time to fall: $t = \sqrt{2h/g} = \sqrt{2(1.25)/10} = 0.50~\text{s}$. Horizontal range: $x = v_x t = (4.0)(0.50) = 2.0~\text{m}$.

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3. Sample 3difficulty 2/5

   A projectile launched at angle $\theta$ from level ground has range $R$. If launch speed is doubled (same angle, same gravity), the new range is

   • A

     $4R$

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   • B

     $R$

   • C

     $8R$

   • D

     $2R$

   Why

   $R \propto v_0^2$, so doubling $v_0$ → $4R$.

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4. Sample 4difficulty 2/5

   For the launch above, what is the horizontal range?

   • A

     $20~\text{m}$

   • B

     $40~\text{m}$

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   • C

     $30~\text{m}$

   • D

     $80~\text{m}$

   Why

   $R = v_0^2 \sin(2\theta)/g = (20)^2 \sin 90^\circ /10 = 400/10 = 40~\text{m}$.

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5. Sample 5difficulty 2/5

   A projectile follows the trajectory shown (no air resistance). Which statement about point <strong>B</strong> (the peak) is correct?

   • A

     Acceleration is zero at B.

   • B

     The horizontal velocity is zero; the vertical velocity is not.

   • C

     Both the horizontal and vertical velocities are zero.

   • D

     The vertical velocity is zero; the horizontal velocity equals the launch value.

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   Why

   Horizontal velocity is constant throughout (no horizontal force), so at the apex it still equals the launch horizontal component. Vertical velocity passes through zero at the peak. Gravity still acts, so acceleration is not zero.

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