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 of the system before the collision is equal to the total momentum after the collision.
Let's use the following variables:
- m1: mass of the bullet (20.5 g = 0.0205 kg)
- v1: initial velocity of the bullet (what we need to find)
- m2: mass of the block (6254 g = 6.254 kg)
- v2: final velocity of the block (1.64 m/s)
Before the collision, the bullet has momentum (p1) given by:
p1 = m1 * v1
After the collision, both the bullet and the block move together, so their total momentum (p2) is:
p2 = (m1 + m2) * v2
Since momentum is conserved, we can equate these two equations:
m1 * v1 = (m1 + m2) * v2
Now we can substitute the given values and solve for v1:
0.0205 kg * v1 = (0.0205 kg + 6.254 kg) * 1.64 m/s
0.0205 kg * v1 = 6.2745 kg * 1.64 m/s
v1 = (6.2745 kg * 1.64 m/s) / 0.0205 kg
v1 = 505.8 m/s
Therefore, the speed of the bullet before striking the block is approximately 505.8 m/s.