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Andrei [34K]
3 years ago
9

A child on a playground swing is swinging back and forth (one complete oscillation) once every four seconds, as seen by her fath

er standing next to the swing. At the same time, a spaceship hurtles by at a speed close to the speed of light. According to special relativity (and ignoring the Doppler effect for this question), a person on the spaceship finds that the time for one full swing is:
A - less than 4 seconds when the spaceship is approaching the swing and longer than 4 seconds when it is moving away.
B - less than 4 seconds.
C - more than 4 seconds.
D - equal to 4 seconds.
Physics
1 answer:
Sati [7]3 years ago
4 0

Answer:

A

Explanation:

Solution:-

- According to the law of relativity the relative speed between two moving objects is inversely proportional to the the time taken.

- Ignoring Doppler Effect.

- So if the relative speeds of two objects in motion i.e ( swing and spaceship) are positive then the time frame of reference for both object relative to other other decreases. So in other words if spaceship approaches the swing i.e relative velocity is positive then the time period of oscillation observed would be less than actual i.e less than 4 seconds.

- Similarly, if spaceship moves away from the swing i.e relative velocity is negative then the time period of oscillation observed would be more than actual i.e more than 4 seconds.

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The change in the momentum of the car due to collision between car and wall is \fbox{\begin\ -23450\text{ kg.m}/\text{s}\end{minispace}} or \fbox{\begin\\-2.3450 \times {10^4}\,\text{kg.m}/\text{s}\end{minispace}}.

Further Explanation:

Let us consider the car is moving towards the right direction and it collides with the wall. After the collision, the car will bounce back or car will rebound opposite to the direction of its motion i.e., towards left. Therefore, the final velocity of the car is opposite to the direction of the initial velocity of the car.

Given:

The mass of the car is 700\,{\text{kg}}.

The velocity of the car before collision is 29\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}.

The velocity of the car after collision is 4.5\text{ m}/\text{s}.  

Concept:

The momentum of an object is defined as the product of mass of object and the velocity with which the object is moving.

The initial momentum of the car is:

\fbox{\begin\\{p_i}= m{v_i}\end{minispace}}                                   …… (I)

Here, {p_i} is the initial momentum of the car, m is the mass of the car and {v_i} is the initial velocity of the car.

Substitute 700\,{\text{kg}} for m and 29\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} for {v_i} in equation (I).  

\begin{gathered}{p_i} = \left( {700\,{\text{kg}}} \right) \cdot \left( {29\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}} \right) \\ = 20300\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} \\ \end{gathered}

The final momentum of the car is defined as the product of mass of car and the velocity of the car after collision or final velocity of the car.

The final momentum of the car is:

\fbox{\begin\\{p_f} = m{v_f}\end{minispace}}                                …… (II)  

Here, {p_f} is the final momentum and {v_f} is the final velocity.

Substitute 700\,{\text{kg}} for m and -4.5\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} for {v_f} in equation (II).  

\begin{aligned}{p_f}&=\left( {700\,{\text{kg}}} \right)\cdot\left( {- 4.5\,{{\text{m}}\mathord{\left/{\vphantom {{\text{m}}{\text{s}}}} \right.\kern-\nulldelimiterspace}{\text{s}}}}\right)\\&=-3150\,{{{\text{kg}} \cdot {\text{m}}}\mathord{\left/{\vphantom{{{\text{kg}}\cdot {\text{m}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}\\\end{aligned}

The change in the momentum of the car after collision is the difference between the momentum of car before collision and the momentum of car after collision.

The change in momentum of the car is:

\fbox{\begin\Delta p = {p_f} - {p_i}\end{minispace}}                             …… (III)  

Here, \Delta p is the change in the momentum of the car.

Substitute - 3150\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} for {p_f} and 20300\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} for {v_i} in equation (III).

\begin{aligned}\Delta{p}&=-3150\text{ kg}.\text{m}/\text{s}\ -20300\text{ kg}.\text{m}/\text{s}\\&=-23450\text{ kg}.\text{m}/\text{s}\end{aligned}

Thus, the change in the momentum of the car due to collision between car and wall is \fbox{\begin\ -23450\text{ kg.m}/\text{s}\end{minispace}} or\fbox{\begin\\-2.3450 \times {10^4}\,\text{kg.m}/{s}\end{minispace}}.

Learn more:

1. The motion of a body under friction brainly.com/question/4033012/  

2. A ball falling under the acceleration due to gravity brainly.com/question/10934170/

3.Conservation of energy brainly.com/question/3943029/    

Answer Details:

Grade: College

Subject: Physics

Chapter: Kinematics

Keywords:

Change in momentum, collision, initial velocity, final velocity, initial momentum, final momentum,-23450 kgm/s^2, -23450 kgm/s2, -2.3450*10^6 kgm/s^2, -2.3450*10^6 kgm/s2.

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