A lead pellet of mass 10.0g is shot horizontally into a stationary wooden block of mass 100g. The pellet hits the block with an
impact velocity of 250ms^–1. It embeds itself in the block and it does not emerge.
What will be the speed of the block immediately after the pellet is embedded?
What will be the speed of the block immediately after the pellet is embedded?
1 answer:
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Answer:
The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s
Explanation:
The momentum of the particle is related to force by the following equation:
Δp = F · Δt
Where:
Δp = change in momentum = final momentum - initial momentum
F = constant force.
Δt = time interval.
Let´s calculate the x-component of the momentum after the 0.13 s:
final momentum - 8 kg m/s = -7 N · 0.13 s
final momentum = -7 kg m/s² · 0.13 s + 8 kg m/s
final momentum = 7.09 kg m/s
Now let´s calculate the y-component of the momentum vector after the 0.13 s. Since the particle wasn´t moving in the y-direction, the initial momentum in this direction is zero:
final momentum = 5 kg m/s² · 0.13 s
final momentum = 0.65 kg m/s
Then, the mometum vector will be as follows:
p = (7.09 kg m/s, 0.65 kg m/s)
The magnitude of this vector is calculated as follows:
The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s