The velocity of the target and arrow after collision is 6.67m/s
<u>Explanation:</u>
Given:
Mass of arrow, mₐ = 415g
Speed of arrow, vₐ = 68.5m/s
Mass of the target, mₓ = 3.3kg = 3300g
speed of the target, vₓ = -1.1m/s (Because the target moves in opposite direction
Velocity of the target and arrow after collision, vₙ = ?
Applying the conservation of momentum,
mₐvₐ + mₓvₓ = (mₐ+mₓ) vₙ
415 X 68.5 + 3300 X -1.1 = (415+3300) X vₙ
28427.5 - 3630 = 3715 X vₙ
24797.5 = 3715 X vₙ
vₙ = 6.67m/s
Therefore, the velocity of the target and arrow after collision is 6.67m/s
Explanation:
Law of conservation of momentum states that in an isolated system when two objects collide with each other then total momentum before and after the collision is equal.
Thus, we can conclude that the law of conservation of momentum states that the total momentum of interacting objects does not change. This means the total momentum before a collision or explosion is equal to the to momentum after a collision or explosion.
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