A bullet of mass 10.6 grams is fired with a speed of 125 meters per second toward a wood block of mass 187 grams initially at rest on a very smooth (assume frictionless) surface.
What is the change in momentum of the bullet if it becomes embedded in the block?
To find the change in momentum of the bullet when it becomes embedded in the block, we need to calculate the initial and final momenta and then find the difference between them.
The momentum of an object is given by the product of its mass and velocity. The formula for momentum is:
Momentum = mass x velocity
Given:
Mass of the bullet (m1) = 10.6 grams = 0.0106 kg
Velocity of the bullet (v1) = 125 meters per second
Mass of the block (m2) = 187 grams = 0.187 kg
Initial velocity of the block (v2) = 0 meters per second (as it is initially at rest)
To find the initial momentum of the bullet, we use the formula:
Initial momentum of the bullet = m1 x v1
Initial momentum of the bullet = 0.0106 kg x 125 m/s
To find the final momentum of the bullet-block system after they become embedded, we need to consider the conservation of momentum. According to this principle, the total momentum before and after the collision should be the same. As the block is initially at rest, its initial momentum is zero.
Total initial momentum = Initial momentum of the bullet
Total final momentum = Final momentum of the bullet-block system
According to the conservation of momentum:
Total initial momentum = Total final momentum
Final momentum of the bullet-block system = Total initial momentum
To find the final momentum of the bullet-block system, we need to consider that the bullet becomes embedded in the block. This means that the block and the bullet move together after the collision with a common final velocity (v3).
Using the conservation of momentum principle, we can write:
Total initial momentum = Total final momentum
(m1 x v1) + (m2 x v2) = (m1 + m2) x v3
Substituting the given values, we have:
(0.0106 kg x 125 m/s) + (0.187 kg x 0 m/s) = (0.0106 kg + 0.187 kg) x v3
Simplifying the equation:
1.325 kg⋅m/s = 0.1976 kg⋅m/s x v3
Divide both sides of the equation by (0.1976 kg⋅m/s):
v3 = 1.325 kg⋅m/s / 0.1976 kg⋅m/s
v3 ≈ 6.703 m/s (rounded to three decimal places)
The final momentum of the bullet-block system is the product of the combined mass (m1 + m2) and the final velocity (v3):
Final momentum of the bullet-block system = (m1 + m2) x v3
Final momentum of the bullet-block system = (0.0106 kg + 0.187 kg) x 6.703 m/s
Finally, the change in momentum (Δp) is the difference between the final momentum and the initial momentum of the bullet:
Change in momentum = Final momentum of the bullet-block system - Initial momentum of the bullet
Change in momentum = [(0.0106 kg + 0.187 kg) x 6.703 m/s] - (0.0106 kg x 125 m/s)
Therefore, the change in momentum when the bullet becomes embedded in the block is the calculated value based on the given mass and velocity information.