Il A 200 g ball is dropped from a height of 2.0 m, bounces...

Key Concepts

Impulse?Momentum Theorem

This principle states that the impulse applied to an object is equal to its change in momentum. It is used to relate the force applied over a time interval to the change in the object's velocity, which is crucial in collision and rebound problems.

Conservation of Energy

When an object is dropped from a height, its gravitational potential energy is converted to kinetic energy as it falls. Using energy conservation, one can determine the speed of the object immediately before impact and after rebound, which is essential for calculating momentum changes.

Momentum Change

Momentum, the product of mass and velocity, changes when an object collides with a surface. The difference between the momentum immediately before and after impact provides a measure of the total impulse, linking the dynamics of the collision to the forces involved.

Force During Impact

By analyzing the collision time and the impulse associated with the momentum change, the average or maximum force exerted during the impact can be calculated. This concept leverages Newton's second law, which shows that force is the rate of change of momentum, making it key to understanding forces in collision events.

Transcript

00:01 Okay, so we know the impulse can be equal to the change momentum, which can be also equal to m delta b, which is equal to m times three final minus vi.

00:09 M is the mass of the ball.

00:14 We have here is the final speed and vi here is the initial speed.

00:19 And so in order to determine the initial speed, we know the initial speed square over 2g should be equal to delta y1.

00:25 Delta y1 is the initial height here, which is 2 .0 meter.

00:28 Okay? and this is the motion for the initial speed.

00:31 So at first, the ball was dropped and then the ball was doing a free -fall motion.

00:40 So as you can tell, the acceleration of gravity is always pointing downward, okay? the initial speed of the whole motion here should be at zero, okay? which is at top here.

00:52 And vi is actually the speed at the bottom, okay? so therefore we have vi squared is equal to 2g times delta y1, which will give us vi is equal to square root 2g delta y1.

01:07 And while this is the graph for the v final, which is the final speed here.

01:11 So once the ball hit on the ground, it will start rebounding back to the, trying to go back to the initial high.

01:19 But eventually it can because the delta y2 here is equal to 1 .5 meter.

01:24 So therefore, as you can tell, the acceleration of gravity is still pointing downward.

01:30 And the final speed of the motion, i'm not talking about the final speed at the bottom here, i'm talking about the final speed at the top of the arrow here, is equal zero, okay? because eventually it will reach a maximum height and you stay there because the time, i'm sorry, the speed is equal zero, okay? so the final speed of the momentum, which is vf here, is actually at the bottom here, okay, which is on the ground.

02:00 Because this is the final speed once the ball was hit on ground.

02:05 So you start changing the direction, point to all upward.

02:09 As you know, delta y2 is equal to 1 .5 meter.

02:12 So therefore we have v final is equal to square root 2g delta y2.

02:18 And then we have the impulse is equal to m times vf minus vi.

02:23 So therefore we have m times square root 2g data y2 minus negative square root 2g delta y1.

02:29 The reason why i put the next here is because the direction of the motion that is related to the initial speed for the momentum and the final speed for the momentum is opposite...

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