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Chapter 3: Problem 24

The units of momentum can be expressed as (A) \(\mathrm{N} \cdot \mathrm{S}\) (B) \(\sqrt{\mathrm{kg} \cdot \mathrm{J}}\) (C) \(\sqrt{\mathrm{kg} \cdot \mathrm{W} \cdot \mathrm{s}}\) (D) all of the above (E) none of the above

Short Answer

Expert verified
(A) \(\mathrm{N} \cdot \mathrm{S}\) (B) \(\sqrt{\mathrm{kg} \cdot \mathrm{J}}\) (C) \(\sqrt{\mathrm{kg} \cdot \mathrm{W} \cdot \mathrm{s}}\) (D) \(\text{kg} \cdot \text{m}^2\) (E) None of the above Answer: (E) None of the above

Step by step solution

01

Write the formula for momentum

The formula for momentum, denoted as "p", is given by: \(p = m \cdot v\) where "m" is mass (in kg) and "v" is velocity (in m/s).
02

Determine the units for momentum

Now, we will determine the unit for momentum by multiplying the units of mass (kg) and velocity (m/s): Units for momentum = \((\text{kg}) \cdot (\text{m/s})\) Units for momentum = \((\text{kg} \cdot \text{m})/\text{s}\)
03

Compare the units for momentum with given options

Now that we have determined the units for momentum, we will compare it with the given options in the exercise: (A) \(\mathrm{N} \cdot \mathrm{S}\) = \(\text{N} \cdot \text{s}\) (B) \(\sqrt{\mathrm{kg} \cdot \mathrm{J}}\) = \(\sqrt{\text{kg} \cdot \text{J}}\) (C) \(\sqrt{\mathrm{kg} \cdot \mathrm{W} \cdot \mathrm{s}}\) = \(\sqrt{\text{kg} \cdot \text{W} \cdot \text{s}}\) From these options, we can see that none of the given options match the actual unit for momentum, which is \((\text{kg} \cdot \text{m})/\text{s}\).
04

Final answer

Since none of the given options match the actual unit for momentum, the correct answer is: (E) none of the above

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Most popular questions from this chapter

Model rocket engines burn for only a short time before using up all their propellant. Suppose a model rocket is launched from its stand at an angle \(60^{\circ}\) above the horizontal. The mass of the rocket is \(0.25 \mathrm{~kg}\) and its final speed is \(50 \mathrm{~m} / \mathrm{s}\). If the engine burns for \(1.25 \mathrm{~s}\), the impulse it gives to the rocket is most nearly (A) \(15.6 \mathrm{~N} \mathrm{~s}\) (B) \(12.5 \mathrm{~N} \mathrm{~s}\) (C) \(6.25 \mathrm{~N} \mathrm{~s}\) (D) \(\frac{\sqrt{3}}{2} \times 12.5 \mathrm{~N} \mathrm{~s}\) (E) \(12.5 \mathrm{~N}\)

A crate of mass \(m\) sits atop a frictionless ramp that has a height and length \(L\). The ramp has mass \(2 m\), which is concentrated at the lower left corner, and it sits on a frictionless ice pond. The crate is released and slides down the ramp. By the instant that the crate reaches the bottom, how far has the ramp moved and in what direction? (A) It hasn't moved. (B) It has moved to the left a distance \(L / 3\). (C) It has moved to the right a distance \(L / 3\). (D) It has moved to the left a distance \(L / 2\). (E) It has moved to the right a distance \(L / 2\).

A block of mass \(5 \mathrm{~kg}\) hangs by a string from the ceiling. cant copy image. The force the ceiling exerts on the string is most nearly: (A) \(1 \mathrm{~N}\), acting up (B) \(5 \mathrm{~N}\), acting down (C) \(50 \mathrm{~N}\), acting down (D) \(50 \mathrm{~N}\), acting up (E) \(100 \mathrm{~N}\), acting up

A crate of mass \(m\) is released from rest at the top of a stationary ramp of height \(h\), whose length along the curve is \(s\), as shown. Assuming that the ramp is frictionless, in calculating the speed of the crate \(v\) at the bottom of the ramp, one can assume: (A) \(h=v t\) (B) \(m g h=1 / 2 m v^2\) (C) \(s=1 / 2 a t^2\) (D) \(v^2=2 a s\) (E) \(h=1 / 2 g t^2\)

Two objects, \(m\) and \(M\), with initial velocities \(v_0\) and \(U_0\) respectively, undergo a one-dimensional elastic collision. Show that their relative speed is unchanged during the collision. (The relative velocity between \(m\) and \(M\) is defined as \(v-U_{-}\))
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