Answer:
-15 m/s
Explanation:
The computation of the velocity of the 4.0 kg fragment is shown below:
For this question, we use the correlation of the momentum along with horizontal x axis
Given that
Weight of stationary shell = 6 kg
Other two fragments each = 1.0 kg
Angle = 60
Speed = 60 m/s
Based on the above information, the velocity = v is
= -15 m/s
Chemical Potential Energy is released when chemical bonds between atoms are broken (like covalent and ionic) and is released mainly as thermal
<span>Elastic Potential is released when the molecules in the material are allowed to go back to there original form, and is released mainly as kinetic</span>
Answer:
u= 200 m/s
Explanation:
Given that
Mass of bullet ,m= 50 gm
Assume that mass of block ,M= 1.2 kg
Lets take speed of the bullet before collision = u m/s
The speed of the system after collision ,v= 8 m/s
There is no any external force ,that is why linear momentum of the system will be conserve.
Linear momentum ,P = mass x velocity
m u = (M+m)v
0.05 x u = (1.2 + 0.05 ) x 8
u= 200 m/s
Therefore the speed of the bullet just before the collision is 200 m/s.