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3 years ago

Describe the conversions between potential and kinetic energy that occur when you shoot a basketball at a basket

1 answer:

6 0

When you shoot a basketball at a basket, the energy from your hands is transferred into the basketball. When you hold the ball up in preparation to shoot, it has potential energy, meaning that it has stored energy that can turn into kinetic energy. When you shoot it, that potential energy turns into kinetic energy, causing the basketball to move.

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A 100 g ball collides elastically with a 300 g ball that is at rest. If the 100 g ball was traveling

sammy [17]

Answer:

The magnitude of the velocities of the two balls after the collision is 3.1 m/s (each one).

Explanation:

We can find the velocity of the two balls after the collision by conservation of linear momentum and energy:

$P_{1} = P_{2}$

$m_{1}v_{1_{i}} + m_{2}v_{2_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}$

Where:

m₁: is the mass of the ball 1 = 100 g = 0.1 kg

m₂: is the mass of the ball 2 = 300 g = 0.3 kg

$v_{1_{i}}$ : is the initial velocity of the ball 1 = 6.20 m/s

$v_{2_{i}}$ : is the initial velocity of the ball 2 = 0 (it is at rest)

$v_{1_{f}}$ : is the final velocity of the ball 1 =?

$v_{2_{f}}$ : is the initial velocity of the ball 2 =?

$m_{1}v_{1_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}$

$v_{1_{f}} = v_{1_{i}} - \frac{m_{2}v_{2_{f}}}{m_{1}}$ (1)

Now, by conservation of kinetic energy (since they collide elastically):

$\frac{1}{2}m_{1}v_{1_{i}}^{2} = \frac{1}{2}m_{1}v_{1_{f}}^{2} + \frac{1}{2}m_{2}v_{2_{f}}^{2}$

$m_{1}v_{1_{i}}^{2} = m_{1}v_{1_{f}}^{2} + m_{2}v_{2_{f}}^{2}$ (2)

By entering equation (1) into (2) we have:

$m_{1}v_{1_{i}}^{2} = m_{1}(v_{1_{i}} - \frac{m_{2}v_{2_{f}}}{m_{1}})^{2} + m_{2}v_{2_{f}}^{2}$

$0.1 kg*(6.20 m/s)^{2} = 0.1 kg*(6.2 m/s - \frac{0.3 kg*v_{2_{f}}}{0.1 kg})^{2} + 0.3 kg(v_{2_{f}})^{2}$

By solving the above equation for $v_{2_{f}}$ :

$v_{2_{f}} = 3.1 m/s$

Now, $v_{1_{f}}$ can be calculated with equation (1):

$v_{1_{f}} = 6.20 m/s - \frac{0.3 kg*3.1 m/s}{0.1 kg} = -3.1 m/s$

The minus sign of $v_{1_{f}}$ means that the ball 1 (100g) is moving in the negative x-direction after the collision.

Therefore, the magnitude of the velocities of the two balls after the collision is 3.1 m/s (each one).

I hope it helps you!

5 0

3 years ago

Scientists have found ancient rock formations dating to the late Proterozoic eon in South America. These ancient rocks can be fo

Dvinal [7]

Answer: Australia

Explanation:

7 0

4 years ago

Read 2 more answers

A 920 kg cannon fires a 3.5 kg shell at initial acceleration of 95 m/s^2. What is the cannon's recoil force?

Mice21 [21]

Answer:

The cannon recoils with a force of 332.5 N

Explanation:

By Newton's third law Recoil force on cannon = Force in shell.

Force in shell = Mass of shell x Acceleration of shell

Mass of shell = 3.5 kg

Acceleration of shell = 95 m/s²

Force in shell = 3.5 x 95 = 332.5 N

Recoil force on cannon = 332.5 N

So, the cannon recoils with a force of 332.5 N

3 0

3 years ago

Which wave has a disturbance parallel to the wave motion

dem82 [27]

Answer:Surface wave

Explanation:

Longitudinal wave is perpendicular to the wave motion and transverse wave is the same as the wave motion

7 0

3 years ago

A proton enters a uniform magnetic field of strength 1 T at 300 m/s. The magnetic field is oriented perpendicular to the proton’

scZoUnD [109]

A charged particle moving in a magnetic field experiences a force equal to:

$\vec{F}=q\vec{v}\times \vec{B}$

Thus, the magnitude of the force that the proton experiences is given by:

$F=qvBsin\theta$

The magnetic field is perpendicular to the proton's velocity, therefore, we have $\theta=90^\circ$ . Replacing the given values, we obtain:

$F=1.6*10^{-19}C(300\frac{m}{s})(1T)sin(90^\circ)\\F=4.8*10^{-17}N$

3 0

3 years ago