Grade 11 Projectile and Equilibrium Physics Practice Quiz (19 questions)

All questions and answers

1. Which feature distinguishes a velocity vector from speed?

Velocity includes both magnitude and direction
Velocity measures magnitude while speed gives direction
Velocity and speed both contain magnitude and direction
Velocity describes direction without a magnitude

Velocity includes both magnitude and direction

Explanation

A velocity vector specifies how fast an object moves and in which direction. Speed is a scalar that gives magnitude without directional information.

2. A velocity has magnitude 20 m/s20\,\text{m/s} at an angle of 30∘30^\circ above the horizontal; what are its components?

vx=20 m/sv_x=20\,\text{m/s} and vy=30 m/sv_y=30\,\text{m/s}
vx=17.3 m/sv_x=17.3\,\text{m/s} and vy=10 m/sv_y=10\,\text{m/s}
vx=10 m/sv_x=10\,\text{m/s} and vy=17.3 m/sv_y=17.3\,\text{m/s}
vx=14.1 m/sv_x=14.1\,\text{m/s} and vy=14.1 m/sv_y=14.1\,\text{m/s}

$$v_x=17.3\,\text{m/s}$$ and $$v_y=10\,\text{m/s}$$

Explanation

Because the angle is measured from the horizontal, the horizontal component is vx=vcos⁡θv_x=v\cos\theta and the vertical component is vy=vsin⁡θv_y=v\sin\theta, giving approximately 17.3 m/s17.3\,\text{m/s} and 10 m/s10\,\text{m/s}. Using sine for the horizontal component would interchange these values.

3. How is the velocity of object B relative to object A calculated?

v⃗B/A=v⃗B−v⃗A\vec v_{B/A}=\vec v_B-\vec v_A
v⃗B/A=v⃗B+v⃗A\vec v_{B/A}=\vec v_B+\vec v_A
v⃗B/A=∣v⃗B∣+∣v⃗A∣\vec v_{B/A}=\left|\vec v_B\right|+\left|\vec v_A\right|
v⃗B/A=v⃗A−v⃗B\vec v_{B/A}=\vec v_A-\vec v_B

$$\vec v_{B/A}=\vec v_B-\vec v_A$$

Explanation

Relative velocity is found by subtracting the reference object's velocity from the observed object's velocity, so v⃗B/A=v⃗B−v⃗A\vec v_{B/A}=\vec v_B-\vec v_A. Adding the velocities does not represent the motion of B as seen from A.

4. A boat points perpendicular to a riverbank while its speed relative to the water stays fixed; what happens if the current becomes faster?

Both crossing time and downstream drift decrease
Both crossing time and downstream drift remain fixed
The crossing time stays fixed while downstream drift increases
The crossing time increases while drift stays fixed

The crossing time stays fixed while downstream drift increases

Explanation

The perpendicular boat component determines the time needed to cross, while the current determines downstream drift. With the perpendicular speed fixed, a faster current increases drift but does not change crossing time.

5. A boat has perpendicular velocity components of 3 m/s3\,\text{m/s} and 4 m/s4\,\text{m/s}; what is its resultant speed?

5 m/s5\,\text{m/s}
7 m/s7\,\text{m/s}
1 m/s1\,\text{m/s}
12 m/s12\,\text{m/s}

$$5\,\text{m/s}$$

Explanation

Perpendicular velocities combine through the square-root sum of squares: v=32+42=5 m/sv=\sqrt{3^2+4^2}=5\,\text{m/s}. Direct addition would apply to collinear velocities rather than perpendicular components.

6. A boat crosses a river with no downstream drift when its upstream velocity component satisfies which condition?

vcos⁡θ=vcv\cos\theta=v_c
vtan⁡θ=vcv\tan\theta=v_c
vsin⁡θ=1vcv\sin\theta=\frac{1}{v_c}
vsin⁡θ=vcv\sin\theta=v_c

$$v\sin\theta=v_c$$

Explanation

No downstream drift occurs when the upstream component balances the current, expressed as vsin⁡θ=vcv\sin\theta=v_c. The cosine component corresponds to the other directional component under the stated angle convention.

7. What initial velocity characterizes a horizontal projectile?

A nonzero vertical velocity directed upward
A vertical velocity and zero horizontal velocity
Equal horizontal and vertical velocity components
A horizontal velocity and zero vertical velocity

A horizontal velocity and zero vertical velocity

Explanation

A horizontal projectile is launched with horizontal velocity while its initial vertical velocity is zero. An angled launch has a nonzero initial vertical component, so it represents a different type of projectile motion.

8. A horizontal projectile is launched from a height of 20 m20\,\text{m} with g=10\,\text{m/s^2}; how long does it remain in flight?

2 s2\,\text{s}
10 s10\,\text{s}
1 s1\,\text{s}
4 s4\,\text{s}

$$2\,\text{s}$$

Explanation

The flight time is determined by the vertical drop: t=2hg=4010=2 st=\sqrt{\frac{2h}{g}}=\sqrt{\frac{40}{10}}=2\,\text{s}. Horizontal launch speed affects range but not the time required to fall from a given height.

9. A projectile is launched at speed v0v_0 and angle θ\theta on level ground. Which expression gives its total flight time?

2v0sin⁡θg\frac{2v_0\sin\theta}{g}
v0sin⁡θ2g\frac{v_0\sin\theta}{2g}
2v0cos⁡θg\frac{2v_0\cos\theta}{g}
v0cos⁡θ2g\frac{v_0\cos\theta}{2g}

$$\frac{2v_0\sin\theta}{g}$$

Explanation

The total flight time depends on the initial vertical component, v0sin⁡θv_0\sin\theta, and is 2v0sin⁡θg\frac{2v_0\sin\theta}{g}. The horizontal component, v0cos⁡θv_0\cos\theta, determines horizontal motion and range rather than flight time.

10. A projectile is launched at speed v0v_0 and angle θ\theta above level ground. How long does it take to reach maximum height?

v0sin⁡θg\frac{v_0\sin\theta}{g}
v0cos⁡θg\frac{v_0\cos\theta}{g}
v0sin⁡θ2g\frac{v_0\sin\theta}{2g}
2v0sin⁡θg\frac{2v_0\sin\theta}{g}

$$\frac{v_0\sin\theta}{g}$$

Explanation

At maximum height, the initial vertical velocity has been reduced to zero by downward acceleration, giving tmax=v0sin⁡θgt_{max}=\frac{v_0\sin\theta}{g}. The acceleration remains downward with magnitude gg even when the vertical velocity is zero.

11. For a projectile launched on level ground with speed v0v_0 at angle θ\theta, which expression gives its maximum height?

v02cos⁡2θ2g\frac{v_0^2\cos^2\theta}{2g}
v02sin⁡2θ2g\frac{v_0^2\sin^2\theta}{2g}
2v02sin⁡2θg\frac{2v_0^2\sin^2\theta}{g}
v0sin⁡θ2g\frac{v_0\sin\theta}{2g}

$$\frac{v_0^2\sin^2\theta}{2g}$$

Explanation

The maximum height is determined by the initial vertical component and equals H=v02sin⁡2θ2gH=\frac{v_0^2\sin^2\theta}{2g}. Using the horizontal component would describe horizontal motion rather than vertical rise.

12. For a fixed launch speed on level ground, which launch angle produces the greatest horizontal range?

45°
60°
90°
30°

45°

Explanation

For a fixed initial speed and equal launch and landing heights, the maximum range occurs at a launch angle of 45°. Angles such as 30° and 60° can produce equal ranges to each other, but not the greatest range.

13. What does the moment of a force describe about a pivot?

The pressure exerted by the force
The turning effect of the force
The mass carried by the force
The speed gained from the force

The turning effect of the force

Explanation

The moment of a force describes its tendency to produce rotation about a pivot. Force by itself measures an interaction, not the rotational effect associated with a particular pivot.

14. A force of 12 N12\,\mathrm{N} acts with a perpendicular distance of 0.50 m0.50\,\mathrm{m} from a pivot. What moment does it produce?

12.5 N m12.5\,\mathrm{N\,m}
24 N m24\,\mathrm{N\,m}
0.042 N m0.042\,\mathrm{N\,m}
6.0 N m6.0\,\mathrm{N\,m}

$$6.0\,\mathrm{N\,m}$$

Explanation

Using τ=Fd\tau=Fd gives τ=(12 N)(0.50 m)=6.0 N m\tau=(12\,\mathrm{N})(0.50\,\mathrm{m})=6.0\,\mathrm{N\,m}. The relevant distance is the perpendicular distance to the line of action, not a different length measured along the lever.

15. A force FF acts at angle θ\theta to a lever of length LL. Which expression gives its moment about the pivot?

FLsin⁡θFL\sin\theta
FLsin⁡θ\frac{FL}{\sin\theta}
FLcos⁡θFL\cos\theta
Fsin⁡θL\frac{F\sin\theta}{L}

$$FL\sin\theta$$

Explanation

The moment is τ=FLsin⁡θ\tau=FL\sin\theta because the perpendicular component of the force produces the turning effect. The cosine expression would correspond to selecting the wrong component for the stated angle.

16. What is the moment about a pivot of a force whose line of action passes through that pivot?

Equal to the lever length
Maximum for that force
Equal to the force magnitude
Zero

Zero

Explanation

A force whose line of action passes through the pivot has zero perpendicular distance, so τ=Fd=0\tau=Fd=0. A nonzero moment requires a perpendicular separation between the pivot and the line of action.

17. When is a body in translational equilibrium?

When its speed is increasing uniformly
When one force acts through its center
When the vector sum of its forces is zero
When the algebraic sum of its moments is zero

When the vector sum of its forces is zero

Explanation

Translational equilibrium requires the vector sum of all forces to be zero, so there is no linear acceleration. A zero moment sum describes rotational equilibrium rather than translational equilibrium.

18. When is a body in rotational equilibrium about a pivot?

When its angular speed is increasing steadily
When the vector sum of its forces is zero
When the algebraic sum of its moments is zero
When its weight acts through the pivot

When the algebraic sum of its moments is zero

Explanation

Rotational equilibrium occurs when clockwise and counterclockwise moments balance so that the algebraic moment sum is zero. A zero net force instead establishes translational equilibrium and does not by itself balance rotation.

19. Which equation represents translational equilibrium for a body?

∑F⃗=0\sum \vec F=0
∑τ=Iα\sum \tau=I\alpha
∑τ=0\sum \tau=0
∑F⃗=ma⃗\sum \vec F=m\vec a

$$\sum \vec F=0$$

Explanation

For translational equilibrium, the vector sum of all forces satisfies ∑F⃗=0\sum \vec F=0. The moment equations describe rotational behavior, while a nonzero force sum corresponds to linear acceleration.

Study with flashcards

Lock in the answers with 39 flashcards on Grade 11 Projectile and Equilibrium Physics.

What two properties does a velocity vector have?

Magnitude and direction.

What is the formula for the horizontal velocity component at angle θ\theta?

vx=vcos⁡θv_x = v \cos \theta.

What is the formula for the vertical velocity component at angle θ\theta?

vy=vsin⁡θv_y = v \sin \theta.

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