An arrow of mass 500 g was released by an archer, moves with the initial velocity of 100 m/s hits the block of mass of 4.5 kg which is at rest initially is placed on a frictionless floor. Find the final velocity of the block and the arrow.
Question
An arrow of mass 500 g was released by an archer, moves with the initial velocity of 100 m/s hits the block of mass of 4.5 kg which is at rest initially is placed on a frictionless floor. Find the final velocity of the block and the arrow.
Solution
To find the final velocity of the block and the arrow, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
The momentum of an object is given by the product of its mass and velocity. Let's denote the mass of the arrow as m1 (0.5 kg) and the mass of the block as m2 (4.5 kg). The initial velocity of the arrow is v1 (100 m/s), and the initial velocity of the block is v2 (0 m/s).
Before the collision, the total momentum is given by: Initial momentum = (m1 * v1) + (m2 * v2)
After the collision, the final velocity of the arrow is v1' and the final velocity of the block is v2'. The total momentum after the collision is given by: Final momentum = (m1 * v1') + (m2 * v2')
Since the collision is elastic (no energy is lost), the total momentum before the collision is equal to the total momentum after the collision. Therefore, we can set up the following equation:
(m1 * v1) + (m2 * v2) = (m1 * v1') + (m2 * v2')
Plugging in the given values: (0.5 kg * 100 m/s) + (4.5 kg * 0 m/s) = (0.5 kg * v1') + (4.5 kg * v2')
Simplifying the equation: 50 kg m/s = 0.5 kg * v1' + 4.5 kg * v2'
Now, we need to solve for v1' and v2'. Since the block is initially at rest, its final velocity v2' will be the velocity it gains from the arrow. Therefore, v2' can be calculated using the equation:
v2' = (m1 * v1) / m2
Plugging in the given values: v2' = (0.5 kg * 100 m/s) / 4.5 kg
Simplifying the equation: v2' = 11.11 m/s
Now that we have the value of v2', we can substitute it back into the equation to solve for v1':
50 kg m/s = 0.5 kg * v1' + 4.5 kg * 11.11 m/s
Simplifying the equation: 50 kg m/s = 0.5 kg * v1' + 49.995 kg m/s
Rearranging the equation: 0.5 kg * v1' = 50 kg m/s - 49.995 kg m/s
Simplifying the equation: 0.5 kg * v1' = 0.005 kg m/s
Dividing both sides by 0.5 kg: v1' = 0.01 m/s
Therefore, the final velocity of the arrow is 0.01 m/s and the final velocity of the block is 11.11 m/s.
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