If coal was our Primary source it would help the world but population would be higher and it can damage animals,people will be able to drive calm without problems ithe world

3 0

4 years ago

A Ping-Pong ball moving East at a speed of 4 m/s collides with a stationary bowling ball. The Ping-Pong ball bounces back to the

In-s [12.5K]

Answer:

They experience the same magnitude impulse

Explanation:

We have a ping-pong ball colliding with a stationary bowling ball. According to the law of conservation of momentum, we have that the total momentum before and after the collision must be conserved:

$p_i = p_f\\p_p + p_b = p'_p+p'_b$

where

$p_p$ is the initial momentum of the ping-poll ball

$p_b$ is the initial momentum of the bowling ball (which is zero, since the ball is stationary)

$p'_p$ is the final momentum of the ping-poll ball

$p'_f$ is the final momentum of the bowling ball

We can re-arrange the equation as follows

$p_p - p'_p = p_b'-p_b$

$-\Delta p_p = \Delta p_b$

which means

$|\Delta p_p | = |\Delta p_b|$ (1)

so the magnitude of the change in momentum of the ping-pong ball is equal to the magnitude of the change in momentum of the bowling ball.

However, we also know that the magnitude of the impulse on an object is equal to the change of momentum of the object:

$I=\Delta p$ (2)

Therefore, (1)+(2) tells us that the ping-pong ball and the bowling ball experiences the same magnitude impulse:

$|I_p| = |I_b|$

4 0

4 years ago

A 22.0 kg bucket of concrete is connected over a very light frictionless pulley to a 375 N box on the roof of a building as show

Veronika [31]

Answer:

vf = 3.27[m/s]

Explanation:

In order to solve this problem we must analyze each body individually and find the respective equations. The free body diagram of each body (box and bucket) should be made, in the attached image we can see the free body diagrams and the respective equations.

With the first free body diagram, we determine that the tension T should be equal to the product of the mass of the box by the acceleration of this.

With the second free body diagram we determine another equation that relates the tension to the acceleration of the bucket and the mass of the bucket.

Then we equalize the two stress equations and we can clear the acceleration.

a = 3.58 [m/s^2]

As we know that the bucket descends 1.5 [m], this same distance is traveled by the box, as they are connected by the same rope.

$x = \frac{1}{2} *a*t^{2}\\1.5 = \frac{1}{2}*(3.58) *t^{2} \\t = 0.91 [s]$