Asked by Shon
A 10.0g bul let traveling horizontally at 365 m/s strikes and becomes embedded in a 2.25kg block of wood placed on a large table. What will be the speed of the bullet block combination immediately after impact ?
Answers
Answered by
herp_derp
Assuming this is an inelastic collision, you use this equation:
m1v1 + m2v2 = (m1 + m2)vf
In this equation let
m1 = mass of bullet
v1 = initial velocity of bullet
m2 = mass of block
v2 = initial velocity of block
vf = final velocity of the bullet-block combination
Okay, let's plug in what we know.
(0.01 kg)(365 m/s) + (2.25 kg)(0 m/s) = (0.01 + 2.25 kg)vf
Notice how v2 is 0 m/s because the block was initially at rest. Then, solve for the final velocity.
3.65 kg-m/s = (2.26 kg)vf
Thus,
vf = 1.62 m/s
m1v1 + m2v2 = (m1 + m2)vf
In this equation let
m1 = mass of bullet
v1 = initial velocity of bullet
m2 = mass of block
v2 = initial velocity of block
vf = final velocity of the bullet-block combination
Okay, let's plug in what we know.
(0.01 kg)(365 m/s) + (2.25 kg)(0 m/s) = (0.01 + 2.25 kg)vf
Notice how v2 is 0 m/s because the block was initially at rest. Then, solve for the final velocity.
3.65 kg-m/s = (2.26 kg)vf
Thus,
vf = 1.62 m/s
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