Find the linear momentum and kinetic energy of a bullet of mass 4.80e-2 kg moving at a speed of 320 m/s.
I found the linear momentum to be 15.36 by doing m*v but I can't figure out the kinetic energy.
yes on the momentum
Ke = (1/2) m v^2
thanks! I'm studying for a test and this helps me tremendously!
To find the kinetic energy of an object, you can use the equation:
Kinetic energy (KE) = (1/2) * mass * velocity^2
Given the mass of the bullet (m = 4.80e-2 kg) and the velocity (v = 320 m/s), we can substitute these values into the equation:
KE = (1/2) * (4.80e-2 kg) * (320 m/s)^2
First, let's square the velocity:
320 m/s * 320 m/s = 102,400 m^2/s^2
Now, substitute this value back into the equation:
KE = (1/2) * (4.80e-2 kg) * (102,400 m^2/s^2)
To solve this, you need to perform the multiplication followed by the division by 2:
KE = (4.80e-2 kg) * (102,400 m^2/s^2) / 2
Now, calculate the result:
KE = 4.9152e3 kg * m^2/s^2
The kinetic energy of the bullet is approximately 4915.2 J (joules).
So, the linear momentum of the bullet is 15.36 kg·m/s, and its kinetic energy is approximately 4915.2 J.