A 4.15-g bullet is moving horizontally with a velocity of +369 m/s, where the sign + indicates that it is moving to the right (see part a of the drawing). The bullet is approaching two blocks resting on a horizontal frictionless surface. Air resistance is negligible. The bullet passes completely through the first block (an inelastic collision) and embeds itself in the second one, as indicated in part b. Note that both blocks are moving after the collision with the bullet. The mass of the first block is 1192 g, and its velocity is +0.635 m/s after the bullet passes through it. The mass of the second block is 1554 g. (a) What is the velocity of the second block after the bullet imbeds itself? (b) Find the ratio of the total kinetic energy after the collision to that before the collision.
To find the velocity of the second block after the bullet embeds itself, we can use the law of conservation of momentum. According to this law, the total momentum before the collision is equal to the total momentum after the collision.
Momentum before collision = Momentum after collision
Before the collision:
The momentum of the bullet is given by:
Momentum of bullet = mass of bullet x velocity of bullet
= 4.15 g x 369 m/s
= 1531.35 g.m/s
= 1.53135 kg.m/s (converting grams to kilograms)
The momentum of the first block is given by:
Momentum of first block = mass of first block x velocity of first block
= 1192 g x 0.635 m/s
= 757.72 g.m/s
= 0.75772 kg.m/s (converting grams to kilograms)
After the collision:
Since the bullet embeds itself in the second block, the momentum of the second block will be equal to the sum of the momentum of the bullet and the first block.
Therefore, the momentum of the second block is given by:
Momentum of second block = Momentum of bullet + Momentum of first block
Substituting the values, we have:
Momentum of second block = 1.53135 kg.m/s + 0.75772 kg.m/s
Momentum of second block = 2.28907 kg.m/s
To find the velocity of the second block after the bullet embeds itself, we divide the momentum by the mass of the second block:
Velocity of the second block = Momentum of second block / mass of second block
= 2.28907 kg.m/s / 1554 g
= 1.4746 m/s
Therefore, the velocity of the second block after the bullet embeds itself is approximately +1.4746 m/s.
Now, let's move on to part (b) to find the ratio of the total kinetic energy after the collision to that before the collision.
The total kinetic energy before the collision is given by:
Total kinetic energy before collision = 0.5 x (mass of bullet x (velocity of bullet)^2 + mass of first block x (velocity of first block)^2)
Total kinetic energy before collision = 0.5 x (4.15 g x (369 m/s)^2 + 1192 g x (0.635 m/s)^2)
Total kinetic energy before collision = 0.5 x (0.005 kg x 136161 m^2/s^2 + 1.192 kg x 0.403225 m^2/s^2) (converting grams to kilograms)
Total kinetic energy before collision = 34687.17 J
After the collision:
The total kinetic energy after the collision is given by the sum of the kinetic energies of the first block and the second block.
The kinetic energy of the first block is given by:
Kinetic energy of first block = 0.5 x mass of first block x (velocity of first block)^2
= 0.5 x 1192 g x (0.635 m/s)^2
= 0.5 x 1.192 kg x 0.403225 m^2/s^2 (converting grams to kilograms)
The kinetic energy of the second block is given by:
Kinetic energy of second block = 0.5 x mass of second block x (velocity of second block)^2
= 0.5 x 1554 g x (1.4746 m/s)^2
= 0.5 x 1.554 kg x 2.17404212 m^2/s^2 (converting grams to kilograms)
Adding the kinetic energies of both blocks, we have:
Total kinetic energy after collision = Kinetic energy of first block + Kinetic energy of second block
Substituting the values, we get:
Total kinetic energy after collision = 0.5 x 1.192 kg x 0.403225 m^2/s^2 + 0.5 x 1.554 kg x 2.17404212 m^2/s^2
Total kinetic energy after collision = 0.242449296 J + 1.68865675668 J
Total kinetic energy after collision = 1.93110605268 J
Now, we can find the ratio of the total kinetic energy after the collision to that before the collision:
Ratio = Total kinetic energy after collision / Total kinetic energy before collision
Ratio = 1.93110605268 J / 34687.17 J
Ratio ≈ 0.0557
Therefore, the ratio of the total kinetic energy after the collision to that before the collision is approximately 0.0557.