Physics, 30.03.2020 23:08 kannstyles709

A 0.140-kg baseball is dropped from rest from a height of 2.20 m above the ground and experiences negligible air resistance as it falls. It rebounds to a height of 1.60 m. What change in the ball's momentum occurs as it rebounds from the ground?

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Answer from: sbailey0962

The change in momentum is -1.70 N*m

Explanation:

Change in momentum (p) is defined as the rest of momentum after the collision (pa) and momentum before the collision (pb):

$\Delta p = p_a-p_b$

Momentum is defined as mass times velocity, then:

$\Delta p = mv_a-mv_b$ (1)

With m the mass of the baseball, va its velocity after the collision and vb velocity before the collision, that is just after and before the ball touches the ground respectively, then our problem is reduced to find those velocities.

To find both we're going to use conservation of energy, the velocity before the collision is the velocity of the ball after falling 2.20m, then we can use conservation of energy as follows:

K_f+U_f=K_i+U_i

With K the kinetic energy and U the potential energy

$\frac{mv_{f}^2}{2}+mgh_f=\frac{mv_{i}^2}{2}+mgh_i$

because the ball is dropped from rest the initial kinetic energy is zero and because the final height of the ball is zero then final potential energy is zero:

$\frac{mv_{f}^2}{2}=mgh_i$

Solving for vf:

$v_{f}=\sqrt{2gh_i}=\sqrt{2(9.81)(2.20)}=6.57\frac{m}{s}=v_b$

Now we should find the velocity just after the collision in a similar way:

K_f+U_f=K_i+U_i

$\frac{mv_{f}^2}{2}+mgh_f=\frac{mv_{i}^2}{2}+mgh_i$

But now the initial momentum is when the ball just reached the ground (potential energy is zero here because is at level ground) and the final one is when the ball is at 1.60m (kinetic energy here is zero because the ball instantly stops), then:

$mgh_f=\frac{mv_{i}^2}{2}$

solving for vi:

$v_{i}=\sqrt{2gh_f}=\sqrt{2(9.81)(1.60)}=5.60\frac{m}{s}=v_a$

It's important to note that because we're working with velocities is important its direction, if we choose negative direction downward then the velocity before the ball touched the ground is positive and the velocity of the ball after touched the ground is negative because is upward, then:

$v_b=6.57\frac{m}{s}$

$v_a=-5.60\frac{m}{s}$

Using those values on equation (1):

$\Delta p = (0.140)(-5.60)-(0.140)(6.57)=-1.70 N*m$

Answer from: Quest

graphing motion: mastery test

209be

what is the instantaneous acceleration of

the particle at point b?

0 meters/second?

10 meters/second?

12 meters/second?

velocity imetersisecond

15 meters/second?

time (seconds)

answer: 0 meters/second^2

Answer from: Quest

a) electrical > light, sound, and thermal< < < <

) the mechanical energy of the firefighter moving.< < < < a) electrical > light, sound, and thermal< < < <

c) both a and b.

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