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Grade 10Mechanics

a coin id dropped in a lift .find the time taken by coin to strike the lift if
a.if lift is at rest
b.lift is going down with v
c.lift is going down with constant v
d.lift is accelerating with acceleration a upward with u =0
1(a
2a>g>
e.lift is accelerating with acceleration a down with u =0
1(a
2(a>g)

Profile image of aakanshika kapoor
11 Years agoGrade 10
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1 Answer

Profile image of Askiitians Tutor Team
ApprovedApproved Tutor Answer1 Year ago

To determine the time taken by a coin to strike the lift in various scenarios, we need to analyze the motion of the coin relative to the lift. The key factors include the lift's state (whether it’s at rest, moving, or accelerating) and the effects of gravity. Let’s break down each case step by step.

Scenario A: Lift at Rest

When the lift is at rest, the only force acting on the coin is gravity. The coin will fall freely under the influence of gravity, which accelerates it at approximately 9.81 m/s².

If we denote the height from which the coin is dropped as h, we can use the equation of motion:

  • h = (1/2)gt²

Rearranging this gives:

  • t = √(2h/g)

Thus, the time taken for the coin to strike the lift is determined solely by the height and the acceleration due to gravity.

Scenario B: Lift Moving Down with Velocity v

In this case, the lift is moving downward at a constant velocity v. The coin, once dropped, will still fall under gravity, but the lift is also moving down. The relative motion can be analyzed as follows:

The effective acceleration of the coin relative to the lift is still g, but since the lift is moving down, we can think of the coin as having an initial downward velocity of v (the same as the lift).

Using the equation of motion:

  • h = vt + (1/2)gt²

Solving for t gives:

  • t = (v + √(v² + 2gh)) / g

Scenario C: Lift Moving Down with Constant Velocity

When the lift is moving down with a constant velocity, the situation is similar to the previous case. The coin will still fall under the influence of gravity, and the lift’s constant velocity does not affect the acceleration of the coin.

Thus, the time taken for the coin to strike the lift remains the same as in Scenario B:

  • t = (v + √(v² + 2gh)) / g

Scenario D: Lift Accelerating Upward with Acceleration a

In this scenario, the lift is accelerating upward with an acceleration a. The effective acceleration acting on the coin becomes (g - a) because the lift's upward acceleration opposes the gravitational pull.

Using the equation of motion again:

  • h = (1/2)(g - a)t²

Rearranging gives:

  • t = √(2h / (g - a))

Scenario E: Lift Accelerating Downward with Acceleration a

Here, the lift is accelerating downward, which means the effective acceleration on the coin is (g + a). This is because the lift's downward acceleration adds to the gravitational pull.

Again, using the equation of motion:

  • h = (1/2)(g + a)t²

Rearranging this gives:

  • t = √(2h / (g + a))

Summary of Time Taken

To summarize, the time taken for the coin to strike the lift varies based on the lift's motion:

  • Lift at rest: t = √(2h/g)
  • Lift moving down with velocity v: t = (v + √(v² + 2gh)) / g
  • Lift accelerating upward with acceleration a: t = √(2h / (g - a))
  • Lift accelerating downward with acceleration a: t = √(2h / (g + a))

Understanding these principles allows us to analyze the motion of objects in different frames of reference effectively. If you have any further questions or need clarification on any point, feel free to ask!