All categories

2 years ago

Suppose the collision between the packages is perfectly elastic. To what height does the package of mass m rebound?

1 answer:

7 0

The height, h to which the package of mass m bounces to depends on its initial velocity, v and the acceleration due to gravity, g and is given below:

$h = \frac{v^{2}}{2g}$

<h3>What are perfectly elastic collision?</h3>

Perfectly elastic collisions are collisions in which the momentum as well as the energy of the colliding bodies is conserved.

In perfectly elastic collisions, the sum of momentum before collision is equal to the momentum after collision.

Also, the sum of kinetic energy before collision is equal to the sum of kinetic energy after collision.

Since some of the Kinetic energy is converted to potential energy of the body;

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

$h = \frac{v^{2}}{2g}$

Therefore, the height to which the package m bounces to depends on its initial velocity and the acceleration due to gravity.

Learn more about elastic collisions at: brainly.com/question/7694106

You might be interested in

5 milligrams into quintal

Greeley [361]

Answer:

divide the mass value by 1e+8

4 0

3 years ago

A comet is traveling through space with speed 3.01 ✕ 104 m/s when it encounters an asteroid that was at rest. The comet and the

Tcecarenko [31]

Answer: $8.493(10)^{-3} m/s$

Explanation:

According to the conservation of linear momentum principle, the initial momentum $p_{i}$ (before the collision) must be equal to the final momentum $p_{f}$ (after the collision):

$p_{i}=p_{f}$ (1)

In addition, the initial momentum is:

$p_{i}=m_{1}V_{1}+m_{2}V_{2}$ (2)

Where:

$m_{1}=1.71(10)^{14} kg$ is the mass of the comet

$m_{2}=6.06(10)^{20} kg$ is the mass of the asteroid

$V_{1}=3.01(10)^{4} m/s$ is the velocity of the comet, which is positive

$V_{2}=0 m/s$ is the velocity of the asteroid, since it is at rest

And the final momentum is:

$p_{f}=(m_{1}+m_{2})V_{f}$ (3)

Where:

$V_{f}$ is the final velocity

Then :

$m_{1}V_{1}+m_{2}V_{2}=(m_{1}+m_{2})V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1}V_{1}}{m_{1}+m_{2}}$ (5)

$V_{f}=\frac{(1.71(10)^{14} kg)(3.01(10)^{4} m/s)}{1.71(10)^{14} kg+6.06(10)^{20} kg}$

Finally:

$V_{f}=8.493(10)^{-3} m/s$ This is the final velocity, which is also in the positive direction.

8 0

4 years ago

If a projectile is fired with an initial velocity of v0 meters per second at an angle α above the horizontal and air resistance

lawyer [7]

Answer:

a) The bullet hits the ground 51.02 s after it was fired.

b) 22,092.3 units

c) 3188.8 units

Explanation:

a) Assuming that the level the bullet was fired from is the ground level.

The bullet hits the ground when y = 0

y = (v₀ sin(α))t − (1/2) gt²

v₀ = 500 m/s

α = 30°

y = 0

0 = (500 sin 30) t - 0.5(9.8)t²

4.9t² - 250t = 0

t(4.9t - 250) = 0

t = 0 s or (4.9t - 250) = 0

The t = 0 s indicates that the bullet was indeed fired from the ground level.

The time it eventually hits the ground back

4.9t = 250

t = 51.02 s

The bullet hits the ground 51.02 s after it was fired.

b) The distance from the firing point that the bullet lands.

x = (v₀ cos(α))t

At this horizontal distance, t = 51.02 s

Substituting the parameters

x = (500 cos 30°) × 51.02

x = 22,092.3 units

c) Maximum height attained by the bullet

Maximum height is given by

H = (u² sin² α)/2g

H = (500² sin² 30°)/(2×9.8)

H = 3188.8 units

Hope this Helps!!!

6 0

4 years ago

Scientists on Earth are able to communicate with rovers on Mars through their receivers, which can receive and process signals.

Bas_tet [7]

B electromagnetic waves I think

4 0

3 years ago

Read 2 more answers

An electric current in a wire flows to the West in a magnetic field directed

vredina [299]

Answer: Current in a wire

We can use the same right-hand rule as we did for the moving charges—pointer finger in the direction the current is flowing, middle finger in the direction of the magnetic field, and thumb in the direction the wire is pushed.

Explanation:

7 0

3 years ago