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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 of motion (at rest, moving down, or accelerating) and the acceleration due to gravity. Let's break down each case systematically.

Case A: Lift at Rest

When the lift is at rest, the only force acting on the coin is gravity. The acceleration of the coin is equal to the acceleration due to gravity, which we denote as g (approximately 9.81 m/s²).

If we assume the coin is dropped from a height h, we can use the equation of motion:

  • h = (1/2)gt²

Rearranging gives:

  • t = √(2h/g)

This formula allows us to calculate the time taken for the coin to hit the lift when it is stationary.

Case B: Lift Moving Down with Velocity v

In this scenario, the lift is moving down at a constant velocity v. The coin is still subject to gravitational acceleration g. The relative motion between the coin and the lift can be analyzed as follows:

  • The effective acceleration of the coin relative to the lift is (g - 0) since the lift's velocity does not affect the acceleration.

Using the same equation of motion as before, we find:

  • h = (1/2)(g)t²

Thus, the time taken remains:

  • t = √(2h/g)

Even though the lift is moving down, the time taken for the coin to strike the lift remains the same as when the lift is at rest.

Case C: Lift Moving Down with Constant Velocity

This case is identical to Case B. The coin still falls under the influence of gravity, and the lift's constant downward velocity does not change the time it takes for the coin to reach the lift. Therefore, the time taken is:

  • t = √(2h/g)

Case D: Lift Accelerating Upward with Acceleration a

When the lift accelerates upward with an acceleration a, the effective acceleration of the coin relative to the lift becomes:

  • (g + a)

Using the equation of motion:

  • h = (1/2)(g + a)t²

Rearranging gives:

  • t = √(2h/(g + a))

In this case, the time taken for the coin to strike the lift is reduced due to the upward acceleration of the lift.

Case E: Lift Accelerating Downward with Acceleration a

For this scenario, where the lift accelerates downward, the effective acceleration of the coin relative to the lift is:

  • (g - a)

Using the same equation of motion:

  • h = (1/2)(g - a)t²

Rearranging gives:

  • t = √(2h/(g - a))

Here, the time taken for the coin to strike the lift increases if the downward acceleration of the lift is less than the acceleration due to gravity.

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 constant velocity: t = √(2h/g)
  • Lift accelerating upward: t = √(2h/(g + a))
  • Lift accelerating downward: t = √(2h/(g - a))

Understanding these principles helps in grasping the concepts of relative motion and acceleration in physics. If you have any further questions or need clarification on any point, feel free to ask!