Advertisements
Advertisements
प्रश्न
A person drops a coin. Describe the path of the coin as seen by the person if he is in
- a car moving at constant velocity and
- in a free falling elevator.
Advertisements
उत्तर
When the person drops the coin, the path of the coin as seen by the person is:
(a)
- When a person drops a coin inside a car moving at constant velocity, the coin appears to fall straight down from the person’s point of view.
- This is because both the person and the coin are moving forward with the same velocity as the car.
- There is no additional force acting on the coin in the horizontal direction, so the coin retains its forward motion and lands directly below the point from which it was dropped.
- This situation reflects the principle of relative motion and supports Newton’s first law an object in motion continues in motion unless acted upon by an external force.
(b)
- In a free falling elevator, both the person and the coin are accelerating downward at the same rate due to gravity.
- When the person drops the coin, from their point of view, it appears to hover, float in the air, or fall very slowly.
- The person and the coin are both in free fall together, so there is no relative acceleration.
- This provides the sensation of weightlessness, similar to what astronauts experience in orbit.
APPEARS IN
संबंधित प्रश्न
A man of mass 70 kg stands on a weighing scale in a lift which is moving
- upwards with a uniform speed of 10 m s-1
- downwards with a uniform acceleration of 5 m s–2
- upwards with a uniform acceleration of 5 m s–2. What would be the readings on the scale in each case?
- What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?
Two masses 8 kg and 12 kg are connected at the two ends of a light, inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.
If the tension in the cable supporting an elevator is equal to the weight of the elevator, the elevator may be
(a) going up with increasing speed
(b) going down with increasing speed
(c) going up with uniform speed
(d) going down with uniform speed
In a TV picture tube, electrons are ejected from the cathode with negligible speed and they attain a velocity of 5 × 106 m/s in travelling one centimetre. Assuming straight-line motion, find the constant force exerted on the electrons. The mass of an electron is 9.1 × 10−31 kg.
The force of buoyancy exerted by the atmosphere on a balloon is B in the upward direction and remains constant. The force of air resistance on the balloon acts opposite the direction of velocity and is proportional to it. The balloon carries a mass M and is found to fall to the earth's surface with a constant velocity v. How much mass should be removed from the balloon so that it may rise with a constant velocity v?
Find the acceleration of the 500 g block in the following figure.

Define linear momentum and state its S.I. unit.
Write the mathematical form of Newton's second law of motion. State the conditions if any.
Use Newton's second law of motion to explain the following instance :
A cricketer pulls his hands back while catching a fast moving cricket ball .
A body of mass 5 kg is moving with velocity 2 m s-1. Calculate its linear momentum.
A force of 10 N acts on a body of mass 2 kg for 3 s, initially at rest. Calculate : Change in momentum of the body.
A force acts for 10 s on a stationary body of mass 100 kg, after which the force ceases to act. The body moves through a distance of 100 m in the next 5 s. Calculate: The velocity acquired by the body.
A pebble is dropped freely in a well from its top. It takes 20 s for the pebble to reach the water surface in the well. Taking g = 10 m s-2 and speed of sound = 330 m s-1. Find : The time when echo is heard after the pebble is dropped.
State Newton's second law of motion. Is Newton's first law of motion contained in Newton's second law of motion?
A cricket ball of mass 150 g has an initial velocity `u = (3hati + 4hatj)` m s−1 and a final velocity `v = - (3hati + 4hatj)` m s−1 after being hit. The change in momentum (final momentum-initial momentum) is (in kg m s1)
Why is catching a slow-moving ball easier than catching a fast-moving ball?
What happens when a car brakes to come to a stop?
