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 it struck the wooden 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. Let's assume the initial velocity of the bullet is denoted as "v" (what we want to find), and the final velocity of the bullet and wooden block combined after the collision is 1.64 m/s.
Before the collision:
Momentum of the bullet = mass of bullet * velocity of bullet = 20.5 g * v
Momentum of the block = mass of the block * 0 (since the block is initially at rest)
After the collision:
Momentum of the bullet and block = (mass of bullet + mass of block) * 1.64 m/s
According to the conservation of momentum principle, the total momentum before the collision is equal to the total momentum after the collision. We can set up the equation:
20.5 g * v = (20.5 g + 6254 g) * 1.64 m/s
First, let's convert grams into kilograms (1 kg = 1000 g):
20.5 g = 0.0205 kg
6254 g = 6.254 kg
Now, we can solve for "v":
0.0205 kg * v = (0.0205 kg + 6.254 kg) * 1.64 m/s
0.0205 kg * v = 6.2745 kg * 1.64 m/s
0.0205 kg * v = 10.27018 kg m/s
Now, divide both sides of the equation by 0.0205 kg to solve for "v":
v = 10.27018 kg m/s / 0.0205 kg
v = 500.5146 m/s
So, the speed of the bullet before striking the block is approximately 500.51 m/s.