Answer: 909 m/s
Explanation:
Given
Mass of the bullet, m1 = 0.05 kg
Mass of the wooden block, m2 = 5 kg
Final velocities of the block and bullet, v = 9 m/s
Initial velocity of the bullet v1 = ? m/s
From the question, we would notice that there is just an object (i.e the bullet) moving before the collision. Also, even after the collision between the bullet and wood, the bullet and the wood would move as one object. Thus, we would use the conservation of momentum to solve
m1v1 = (m1 + m2) v, on substituting, we have
0.05 * v1 = (0.05 + 5) * 9
0.05 * v1 = 5.05 * 9
0.05 * v1 = 45.45
v1 = 45.45 / 0.05
v1 = 909 m/s
Thus, the original velocity of the bullet was 909 m/s
The original velocity of the bullet is 909 m/s.
Let [tex]m_{1}[/tex] be the mass of bullet and [tex]m_{2}[/tex] be the mass of wooden block.
And also let V be the final velocity of the bullet and block and [tex]V_{1}[/tex] be the velocity of the bullet.
Given, [tex]m_{1}[/tex] will be 0.05 kg and [tex]m_{2}[/tex] will be 5 kg.
Final velocity V is 9 m/s.
we have to calculate initial velocity of the bullet.
From the principle of conservation of momentum.
[tex]m_{1} v_{1}[/tex] [tex]=[/tex][tex](m_{1} +m_{2} ) V[/tex]
Now putting the values of [tex]m_{1} , m_{2} and V[/tex]
[tex]0.05[/tex][tex]V_{1}[/tex] [tex]= 9 (0.05+5)[/tex]
[tex]0.05V_{1} = 9( 5.05)[/tex]
[tex]0.05V_{1} = 45.45[/tex]
[tex]V_{1} = 909[/tex]
Hence the original velocity of the bullet is 909 m/s.
For more details on conservation of momentum follow the link:
https://brainly.com/question/347616
