- A baseball pitcher throws a pitch at 42 meters per second. If the batter is 18 meters from the pitcher, approximately how much time does it take the ball to reach the batter?
- 1.8 seconds
- 2 seconds
- 0.87 seconds
- 0.43 seconds
Use \(t=\dfrac{d}{v}\):
\(t=\dfrac{18\ \text{m}}{42\ \text{m/s}}=\dfrac{3}{7}\ \text{s}\approx 0.43\ \text{s}\)
So the ball reaches the batter in about 0.43 s.
- How much work is done on a downhill skier by an average braking force of \(9.9 \times 10^2 \text{ N}\) to stop her in a distance of 10 meters?
- \(2 \times 10^2 \text{J}\)
- \(3.5 \times 10^2 \text{J}\)
- \(8.4 \times 10^3 \text{J}\)
- \(9.9\times 10^3\ \text{J}\)
The braking force points opposite the motion, so \(\theta=180°\) and \(\cos 180°=-1\).
\(W=(9.9\times 10^2\ \text{N})(10\ \text{m})(-1)\) \(=-9.9\times 10^3\ \text{J}\)
The braking force does negative work (it removes kinetic energy), so the magnitude is \(9.9\times 10^3\ \text{J}\).
- A bicyclist accelerates from rest to a speed of 5 meters per second in 10 seconds. During the same 10 seconds, a car accelerates from a speed of 22 meters per second to a speed of 27 meters per second. Compared to the acceleration of the bicycle, what is the acceleration of the car?
- The same
- Greater
- Less
- Unknown
Acceleration is \(a=\dfrac{\Delta v}{\Delta t}\).
Knowing this, we can calculate the acceleration of the bicycle first:
\(a_b=\dfrac{5-0}{10}=0.5\ \text{m/s}^2\)
Then, we can calculate the acceleration of the car and compare the two:
\(a_c=\dfrac{27-22}{10}=0.5\ \text{m/s}^2\)
Therefore, the acceleration for the bicycle and the car is exactly the same.
- A baseball player throws a baseball at a speed of 40 meters per second at an angle of 30 degrees. What is the horizontal component of the baseball’s speed?
- 20 m/s
- 25 m/s
- 30 m/s
- 35 m/s
To find the horizontal component of the ball’s speed, multiply the launch speed by the cosine of the angle:
\(v_x=v\cos\theta=40\cos 30°\) \(=40\left(\dfrac{\sqrt{3}}{2}\right)\) \(\approx 34.6\ \text{m/s}\)
Rounding gives about 35 m/s, so the horizontal component is 35 m/s.
- What is the average velocity of a car that travels due west at 30 kilometers in 0.5 hr?
- 60 km/hr west
- 60 km/hr
- 15 km/hr west
- 15 km/hr
Average velocity equals displacement divided by time, with direction included. Here, \(\bar v=\frac{30\ \text{km}}{0.5\ \text{hr}}\) \(=60\ \text{km/hr}\), and because the car travels due west, the average velocity is 60 km/hr west.
- A man weighs 900 N standing on a scale in a stationary elevator. If some time later the reading on the scale is 1,200 N, the elevator must be moving with…
- Constant acceleration downward
- Constant speed downward
- Constant acceleration upward
- Constant speed upward
A scale reads the normal force, not the weight itself. When the reading rises from 900 N to 1,200 N, we must have \(N>mg\), which implies an upward net force and therefore an upward acceleration.
Thus, the elevator is moving with constant upward acceleration, not constant speed (which would keep \(N=mg\)).
- Net force \(F\) causes mass \(m_1\) to accelerate at rate \(a_1\). A net force of \(3F\) causes \(m_2\) to accelerate at a rate of \(2a_1\). What is the ratio of \(m_1\) to \(m_2\)?
- 2:3
- 3:4
- 1:2
- 2:1
Using \(F=ma\), the first situation gives \(m_1=\frac{F}{a_1}\).
In the second, \(3F=m_2(2a_1)\), so \(m_2=\frac{3F}{2a_1}\).
Comparing the two masses:
\(\dfrac{m_1}{m_2}=\dfrac{F/a_1}{(3F)/(2a_1)}\) \(=\dfrac{2}{3}\)
Thus, the ratio \(m_1:m_2\) is 2:3.
- An artillery shell is fired at an angle to the horizontal. Its initial velocity has a vertical component of 150 meters per second and a horizontal component of 260 meters per second. What is the magnitude of the initial velocity of the shell?
- \(2.5 \times 10^2 \text{ m/s}\)
- \(3 \times 10^2 \text{ m/s}\)
- \(3.5 \times 10^3 \text{ m/s}\)
- \(3.5 \times 10^4 \text{ m/s}\)
The initial speed is the magnitude of the velocity vector formed by its perpendicular components, so:
\(v_0=\sqrt{v_x^2+v_y^2}\) \(=\sqrt{(260)^2+(150)^2}\) \(=\sqrt{90,100}\approx 3.0\times 10^2\ \text{m/s}\)
Therefore, the shell’s initial speed is about \(3\times 10^2\ \text{m/s}\).
- Into how many possible components can a single force be resolved?
- An unlimited number
- Two components
- Three components
- Four components at right angles to each other
A single force vector can be expressed as the sum of any number of component vectors whose vector sum equals the original. Choosing two perpendicular components is merely convenient, not mandatory. Therefore, a force can be resolved into an unlimited number of components.
- What is the magnitude of the gravitational force between two 5 kg masses separated by a distance of five meters?
- \(1.3 \times 10^{-14} \text{N}\)
- \(3.3 \times 10^{-10} \text{N}\)
- \(6.7 \times 10^{-11} \text{N}\)
- \(1.3 \times 10^{-11} \text{N}\)
Applying Newton’s law of gravitation:
\(F=\dfrac{Gm_1m_2}{r^2}\) \(=\dfrac{(6.67\times 10^{-11})(5)(5)}{(5)^2}\) \(=6.67\times 10^{-11}\ \text{N}\)
Note that the 25 in the numerator and denominator cancel.
- Two cars having different weights are traveling on a level surface at different constant velocities. Within the same time interval, greater force will always be required to stop the car that has greater…
- Weight
- Kinetic energy
- Velocity
- Momentum
Stopping a car in the same time interval requires an average force equal to the change in momentum over time:
\(\bar F=\dfrac{\Delta p}{\Delta t}\)
The car with the greater momentum will therefore require the greater stopping force in the same \(\Delta t\), so the correct criterion is greater momentum.
- A 0.05 kg bullet is fired from a 4 kg rifle that is initially at rest. If the bullet leaves the rifle with momentum having a magnitude of \(20 \text{ kg} \times \text{ m/s}\), the rifle will recoil with a momentum having a magnitude of…
- \(1,600 \text{ kg} \times \text{m/s}\)
- \(80 \text{ kg} \times \text{m/s}\)
- \(20 \text{ kg} \times \text{m/s}\)
- \(0.25 \text{ kg} \times \text{m/s}\)
Because the rifle–bullet system starts at rest, total momentum is zero before firing and must remain zero after firing. This means the rifle’s recoil momentum is equal in magnitude and opposite in direction to the bullet’s momentum, so its magnitude is also \(20\ \text{kg}\times\text{m/s}\).
- A wooden block is at rest on a horizontal steel surface. If a 10 N of force applied parallel to the surface is required to put the block in motion, how much force is required to keep the block moving at a constant velocity?
- <10 N
- >10 N
- 10 N
- Unknown
Starting motion requires overcoming static friction, whereas keeping motion at constant velocity only needs to balance kinetic friction, and typically \(\mu_k\lt \mu_s\).
Since it took 10 N to start the block, less than 10 N will keep it moving at constant speed.
- A girl weighing 500 N takes 50 seconds to climb a flight of stairs 18 meters high. What is her power output vertically?
- 120 W
- 150 W
- 180 W
- 220 W
Her work against gravity is:
\(W=Fd=(500\ \text{N})(18\ \text{m})=9,000\ \text{J}\)
Dividing by the time gives the power \(\frac{W}{t}\) \(=\frac{9,000}{50}\) \(=180\ \text{W}\), so her vertical power output is 180 W.
- The path of a projectile fired at a 30° angle to the horizontal is best described as…
- Parabolic
- Linear
- Circular
- Hyperbolic
With air resistance neglected, the horizontal motion of a projectile is uniform while the vertical motion has constant downward acceleration, and together these produce a parabolic trajectory.
- A projectile is launched with an initial velocity of 20 meters per second at an angle of 30° above the horizontal. What is the magnitude of the vertical component of the projectile’s initial velocity?
- \(20 \text{ m/s} \times \text{cos}30°\)
- \(20 \text{ m/s} \times \text{sin}30°\)
- \(\frac{20 \text{ m/s}}{\text{sin}30°}\)
- \(\frac{20 \text{ m/s}}{\text{cos}30°}\)
The vertical component of the initial velocity is found with sine:
\(v_y=v\sin\theta=20\sin 30°=10\ \text{m/s}\)
Thus, the correct expression is \(20\ \text{m/s}\times \sin 30°\), yielding 10 m/s.