A bullet of mass 35g strikes a stationary wooden block of mass 2.6kg, becomes embedded in the block, and they both move off at 7.5m/s. What was the original velocity of the bullet?
m1 = 35 g = 0.035 kg, v1 =?
m2 = 2.6 kg, v2 =0,
u = 7.5 m/s
m1•v1 =(m1+m2) •u,
v1 =(m1+m2) •u/m1 = ......
564'64m/s
To find the original velocity of the bullet before it collided with the wooden block, we can use the principle of conservation of momentum.
According to the principle of conservation of momentum, the total momentum of a system remains constant before and after a collision, provided no external forces are acting on the system.
The momentum of an object is defined as the product of its mass and velocity. Mathematically, it is expressed as:
Momentum (p) = mass (m) * velocity (v)
Let's represent the initial velocity of the bullet as "u."
Initially, the bullet is moving, and the wooden block is stationary. Therefore, the total momentum before the collision is given by:
Initial momentum = Bullet's momentum (p_bullet) + Block's momentum (p_block)
= (mass of bullet * initial velocity of bullet) + (mass of block * 0)
= mass of bullet * initial velocity of bullet
After the collision, the bullet becomes embedded in the wooden block, and they both move off together at a common velocity (7.5 m/s). So, the total momentum after the collision is:
Final momentum = Total mass of bullet and block * final velocity
= (mass of bullet + mass of block) * final velocity
= (mass of bullet + mass of block) * 7.5 m/s
According to the principle of conservation of momentum, the initial momentum equals the final momentum. So we can equate these two values:
mass of bullet * initial velocity of bullet = (mass of bullet + mass of block) * 7.5 m/s
Now, we can substitute the given masses and solve for the initial velocity of the bullet:
0.035 kg * u = (0.035 kg + 2.6 kg) * 7.5 m/s
Simplifying the equation:
0.035 kg * u = 2.635 kg * 7.5 m/s
Dividing both sides by 0.035 kg:
u = (2.635 kg * 7.5 m/s) / 0.035 kg
Calculating:
u = 565.714 m/s
Therefore, the original velocity of the bullet was approximately 565.714 m/s.