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A metal plate weighing 200 gram is balanced in mid air by throwing 40 balls per second vertically upwards from below after collision balls rebound with the same speed what does the speed with which the ball strike the plate. given mass of each ball = 200 g and g = 10 m/s²?
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A metal plate weighing 200 gram is balanced in mid air by throwing 40 ...
Problem Overview
To balance a metal plate weighing 200 grams using balls thrown upwards, we need to determine the speed at which each ball strikes the plate. Given that the mass of each ball is 200 grams and they are released at a rate of 40 balls per second, we can use the principles of momentum and forces to solve this.

Given Data
- Weight of the metal plate (W) = 200 grams = 0.2 kg
- Mass of each ball (m) = 200 grams = 0.2 kg
- Gravitational acceleration (g) = 10 m/s²
- Rate of balls thrown (N) = 40 balls/second

Calculating the Force Required
- The weight of the plate generates a downward force:
Weight of the plate = W = mg = 0.2 kg × 10 m/s² = 2 N
- To balance this, the upward force from the balls must equal 2 N.

Upward Force from Balls
- Each ball, upon striking the plate with speed \( v \), exerts a force during the collision:
Force from each ball = Change in momentum = mv
- Since balls rebound with the same speed \( v \), the change in momentum for each ball is \( 2mv \) (as it goes up and comes down).
- Therefore, the total force from all balls per second is:
Total force = N × (2mv) = 40 × (2 × 0.2 × v) = 16v

Setting Forces Equal
- To balance the forces:
16v = 2
- Solving for \( v \):
v = 2 / 16 = 0.125 m/s

Conclusion
Thus, the speed at which each ball strikes the plate is **0.125 m/s**.
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A metal plate weighing 200 gram is balanced in mid air by throwing 40 balls per second vertically upwards from below after collision balls rebound with the same speed what does the speed with which the ball strike the plate. given mass of each ball = 200 g and g = 10 m/s²?
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