A miniature spring-loaded, radio-controlled gun is mounted on an air puck. The gun's bullet has a mass of 5.00 g, and the gun and puck have a combined mass of 120 g. With the system initially at rest, the radio controlled trigger releases the bullet causing the puck and empty gun to move with a speed of 0.500 m/s. What is the bullet's speed?

momentum is conserved

the momentum of the bullet is equal to the momentum of the puck/gun
... but in the opposite direction

(mass bullet) * (speed bullet) = (mass puck/gun) * (speed puck/gun)

To find the bullet's speed, we can use the principle of conservation of momentum.

The initial momentum of the system (bullet + gun + puck) is zero, as the system is at rest.

The final momentum of the system can be calculated using the formula:

Final Momentum = (Mass of bullet + Mass of gun + Mass of puck) × Final Velocity

Let's denote the mass of the bullet as "m₁" and the mass of the gun and puck as "M₂". The final velocity of the system is given as 0.500 m/s.

Final Momentum = (m₁ + M₂) × 0.500

Since the bullet is released, its momentum is only due to its own mass and velocity, so the final momentum of the bullet is given by:

Bullet's Final Momentum = m₁ × Bullet's Final Velocity

To conserve momentum, the sum of the final momenta of the bullet and the gun+puck system should be equal to zero:

(m₁ + M₂) × 0.500 + m₁ × Bullet's Final Velocity = 0

We are given the value of M₂ as 120 g, which is equal to 0.120 kg.

Substituting the known values into the equation, we have:

(5.00 × 10⁻³ + 0.120) × 0.500 + 5.00 × 10⁻³ × Bullet's Final Velocity = 0

Simplifying the equation further, we get:

(0.125 + 0.120) + 0.005 × Bullet's Final Velocity = 0

0.245 + 0.005 × Bullet's Final Velocity = 0

0.005 × Bullet's Final Velocity = -0.245

Dividing by 0.005:

Bullet's Final Velocity = -0.245 ÷ 0.005

Bullet's Final Velocity = -49 m/s

Since velocity cannot be negative in this case (as it represents speed), we discard the negative value and take the magnitude:

Bullet's Final Velocity = 49 m/s

Therefore, the bullet's speed is 49 m/s.

To solve this problem, we need to apply the law of conservation of momentum. The law of conservation of momentum states that the total momentum of an isolated system remains constant if no external forces act on it.

In this case, the system consists of the bullet, gun, and puck. Initially, the system is at rest, so the initial momentum is zero. After the bullet is released, the momentum is still conserved, but now shared between the gun, puck, and bullet.

Let's denote the bullet's speed as v, the gun and puck's combined mass as M, and the puck's speed as V.

According to the law of conservation of momentum:

Initial momentum = Final momentum

(0) = (M + m)Vfinal

Where m is the mass of the bullet.

The final momentum of the system is the momentum of the gun and puck, which can be calculated as:

Momentum of gun and puck = (M + m)Vfinal

Since the bullet is released and moves with a speed, we can calculate its momentum as:

Momentum of the bullet = m * v

Now, since momentum is conserved, we can equate the two equations:

(M + m)Vfinal = m * v

Now we can substitute the given values into the equation:

(120 g + 5.00 g)(0.500 m/s) = 5.00 g * v

Converting all masses to kilograms:

(0.125 kg + 0.005 kg)(0.500 m/s) = 0.005 kg * v

Now, we can solve for v by dividing both sides of the equation by 0.005 kg:

v = (0.130 kg * m/s) / 0.005 kg

v = 26 m/s

Therefore, the bullet's speed is 26 m/s.

12 J