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svetoff [14.1K]

3 years ago

6

A 1.0-kg ball has a velocity of 12 m/s downward just before it strikes the ground and bounces up with a velocity of 12 m/s upwar

d. What is the change in momentum of the ball?

1 answer:

Nezavi [6.7K]3 years ago

3 0

Answer:

The change in momentum of the ball is 24 kg-m/s

Explanation:

It is given that,

Mass of the ball, m = 1 kg

Initial velocity of the ball, u = -12 m/s (in downwards)

Final velocity of the ball, v = +12 m/s (in upward)

We need to find the change in momentum of the ball.

Initial momentum of the ball, $p_i=mu=1\ kg\times (-12\ m/s)=-12\ kg-m/s$

Final momentum of the ball, $p_f=mv=1\ kg\times (12\ m/s)=12\ kg-m/s$

Change in momentum of the ball, $\Delta p=p_f-p_i$

$\Delta p=12-(-12)=24\ kg-m/s$

So, the change in momentum of the ball is 24 kg-m/s. Hence, this is the required solution.

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A rocket rises vertically, from rest, with an acceleration of 3.6m/s^2 until it runs out of fuel at an altitude of 1500m. After

Montano1993 [528]

A. 103.9 m/s

The motion of the rocket until it runs out of fuel is an accelerated motion with constant acceleration a = 3.6 m/s^2, so we can use the following equation

$v^2 -u^2 = 2ad$

where

v is the velocity of the rocket when it runs out of fuel

u = 0 is the initial velocity of the rocket

a = 3.6 m/s^2 is the acceleration

d = 1500 m is the distance covered during this first part

Solving for v, we find

$v=\sqrt{u^2+2ad}=\sqrt{(0^2+2(3.6 m/s^2)(1500 m)}=103.9 m/s$

B. 28.9 s

We can calculate the time taken for the rocket to reach this altitude with the formula

$d=\frac{1}{2}at^2$

where

d = 1500 m

a = 3.6 m/s^2

Solving for t, we find

$t=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2(1500 m)}{3.6 m/s^2}}=28.9 s$

C. 2050.8 m

We can calculate the maximum altitude reached by the rocket by using the law of conservation of energy. In fact, from the point it runs out of fuel (1500 m above the ground), the rocket experiences the acceleration due to gravity only, so all its kinetic energy at that point is then converted into gravitational potential energy at the point of maximum altitude:

$K_i = U_f\\\frac{1}{2}mu^2 = mgh$

where h is distance covered by the rocket after it runs out of fuel, and v=103.9 m/s is the velocity of the rocket when it starts to decelerate due to gravity. Solving for h,

$h=\frac{v^2}{2g}=\frac{(103.9 m/s)^2}{2(9.8 m/s^2)}=550.8 m$

So the maximum altitude reached by the rocket is

$h' = d+h=1500 m +550.8 m=2050.8 m$

D. 39.5 s

The time needed for the second part of the trip (after the rocket has run out of fuel) can be calculated by

$h=\frac{1}{2}gt^2$

where

h = 550.8 m is the distance covered in the second part of the trip

g = 9.8 m/s^2

Solving for t,

$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(550.8 m)}{9.8 m/s^2}}=10.6 s$

So the total time of the trip is

$t'=28.9 s+10.6 s=39.5 s$

E. 200.5 m/s

When the rocket starts moving downward, it is affected by gravity only. So the gravitational potential energy at the point of maximum altitude is all converted into kinetic energy at the instant the rocket hits the ground:

$\frac{1}{2}mv^2 = mgh$

where

v is the final velocity of the rocket

h = 2050.8 m is the initial altitude of the rocket

Solving for v,

$v=\sqrt{2gh}=\sqrt{2(9.8 m/s^2)(2050.8 m)}=200.5 m/s$

F. 60 s

We need to calculate the time the rocket takes to fall down to the ground from the point of maximum altitude, and that is given by

$h'=\frac{1}{2}gt^2$

where

h' = 2050.8 m

g = 9.8 m/s^2

Solving for t,

$t=\sqrt{\frac{2h'}{g}}=\sqrt{\frac{2(2050.8 m)}{9.8 m/s^2}}=20.5 s$

So the total time of the trip is

$t''=39.5 s+20.5 s=60 s$

6 0

4 years ago

A baseball with a mass of 0.125 kg is hit toward the pitcher at a speed of 45 m/s. The pitcher's glove stops the baseball in 0.0

Vladimir [108]

It would be the first one

3 0

3 years ago

Read 2 more answers

What is HALF PROJECTILE MOTION and what are some examples of it?

antiseptic1488 [7]

I think basically anything you throw vertically, from that point to its maximum height where you get a half semi circle.. or if you throw a ball horizontally, from that point to the point it comes to rest.

5 0

3 years ago

If the car moves for equal times along the road and hill, create an expression for its average velocity vector v(ave) in terms o

SIZIF [17.4K]

So here in order to find the average velocity we can say

$v_{avg} = \frac{displacement}{time}$

so first we know that along the horizontal and along the inclined it moves with same time interval

so here we will have displacement in x direction as

$x = v_o (t) + v_{1x}(t)$

now the average velocity in x direction will be given as

$v_{avg} = \frac{v_o t + v_{1x} t}{t + t}$

$v_{avg} = \frac{v_o + v_{1x}}{2}$

now similarly for y direction

first we will find its displacement

$y = v_{1y}(t)$

now the average velocity in y direction will be given as

$v_{avg} = \frac{v_{1y} t}{t + t}$

$v_{avg} = \frac{v_{1y}}{2}$

now net velocity is given as

$v_{avg} = \frac{v_o + v_{1x}}{2}\hat i + \frac{v_{1y}}{2}\hat j$

5 0

3 years ago

A planet is 10 light years away from Earth. What speed would you need to go for a trip to the planet and back to take only 5 yea

viva [34]

Answer:

a. speed, v = 0.97 c

b. time, t' = 20.56 years

Given:

t' = 5 years

distance of the planet from the earth, d = 10 light years = 10 c

Solution:

(a) Distance travelled in a round trip, d' = 2d = 20 c = L'

Now, using Length contraction formula of relativity theory:

$L'' = L'\sqrt{1 - \frac{v^{2}}{c^{2}}}$ (1)

time taken = 5 years

We know that :

time = $\frac{distance}{speed}$

5 = $\frac{L''}{v}$ (2)

Dividing eqn (1) by v on both the sides and substituting eqn (2) in eqn (1):

$\frac{L'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5$

$\frac{20'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5$

Squaring both the sides and Solving above eqution, we get:

v = 0.97 c

(b) Time observed from Earth:

Using time dilation:

$t'' = \frac{t'}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$

$t'' = \frac{5}{\sqrt{1 - \frac{(0.97c)^{2}}{c^{2}}}}$

Solving the above eqn:

t'' = 20.56 years

4 0

3 years ago