A(n) 20.5 g bullet is shot into a(n) 6254 g wooden block standing on a frictionless surface.The block, with the bullet in it, acquires a speed of 1.64 m/s.
Calculate the speed of the bullet before striking the block.
Thank you
To calculate the speed of the bullet before striking the block, 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:
Momentum = Mass × Velocity
Let's assume the initial velocity of the bullet before striking the block is v. And let's label the mass of the block as M and the mass of the bullet as m.
According to the conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision:
(mass of bullet × initial velocity of bullet) + (mass of block × initial velocity of block) = (mass of bullet × final velocity of bullet inside the block) + (mass of block × final velocity of block)
Given:
mass of bullet (m) = 20.5 g = 0.0205 kg
mass of block (M) = 6254 g = 6.254 kg
final velocity of block (v_block) = 1.64 m/s
We need to solve for the initial velocity of the bullet (v_bullet).
Using the conservation of momentum equation:
(0.0205 kg × v_bullet) + (6.254 kg × 0) = (0.0205 kg × 1.64 m/s) + (6.254 kg × 1.64 m/s)
Since the block is initially at rest (velocity = 0), the second term on the left side becomes 0.
(0.0205 kg × v_bullet) = (0.0205 kg × 1.64 m/s) + (6.254 kg × 1.64 m/s)
Multiplying through:
0.0205 kg × v_bullet = 0.03362 kg·m/s + 10.24456 kg·m/s
Combining the two terms:
0.0205 kg × v_bullet = 10.27818 kg·m/s
Dividing by the mass of the bullet:
v_bullet = 10.27818 kg·m/s / 0.0205 kg
v_bullet ≈ 501.9 m/s
Therefore, the speed of the bullet before striking the block is approximately 501.9 m/s.