A 7.45-g bullet is moving horizontally with a velocity of +348 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 1220 g, and its velocity is +0.648 m/s after the bullet passes through it. The mass of the second block is 1603 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.
m=7,45 g, m1=1220 g, m2=1603 g,
v=348 m/s, v1 = 0.648 m/s
The law of conservation of linear momentum for the 1st collision
m•v+0=m1•v1+m•u
u=(m•v-m1•v1)/m = v –(m1/m)v1=
348 – (1220/7.45) •0.648 = 241.9 m/s,
The law of conservation of linear momentum for the 2nd collision
mu+0=(m+m1)u1
u1= mu/(m1+m) = 7.45•241.9/(1603+7.45) =1.12 m/s.
Before the 1st collision KE= m•v²/2,
After the 1st collision
KE1=m1•v1²/2+m•u²/2
After the 2nd collision
KE2=(m+m1) •u1²/2
To solve this problem, we can use the law of conservation of momentum and the law of conservation of energy. Let's break down the steps to find the answers to both parts of the question:
Step 1: Find the velocity of the first block after the collision
- The initial momentum before the collision is the same as the momentum after the collision.
- The momentum of the bullet before the collision is given by (mass of bullet) × (velocity of bullet).
- The momentum of the first block after the collision is given by (mass of first block) × (velocity of first block).
- Set up the equation: (mass of bullet) × (velocity of bullet) = (mass of first block) × (velocity of first block).
- Substitute the given values: (7.45 g) × (348 m/s) = (1220 g) × (0.648 m/s).
- Calculate the velocity of the first block after the collision.
Step 2: Find the velocity of the second block after the collision
- The momentum of the second block is given by (mass of second block) × (velocity of second block).
- Since the bullet embeds itself in the second block, the momentum of the second block is the sum of the bullet's momentum and the first block's momentum.
- Set up the equation: (mass of second block) × (velocity of second block) = (momentum of bullet) + (momentum of first block).
- Substitute the given values: (1603 g) × (velocity of second block) = (7.45 g) × (348 m/s) + (1220 g) × (velocity of first block).
- Plug in the known values and the velocity of the first block calculated in Step 1.
- Calculate the velocity of the second block after the collision.
Step 3: 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 (1/2) × (mass of bullet) × (velocity of bullet)^2.
- The total kinetic energy after the collision is given by (1/2) × (mass of first block + mass of second block) × (velocity of second block)^2.
- Set up the equation: (total kinetic energy after the collision) / (total kinetic energy before the collision) = [(1/2) × (mass of first block + mass of second block) × (velocity of second block)^2] / [(1/2) × (mass of bullet) × (velocity of bullet)^2].
- Substitute the known values and the velocity of the first block calculated in Step 1.
- Calculate the ratio of the total kinetic energy after the collision to that before the collision.
Please note that this is just a general outline of the steps involved. To get the actual numerical answers, you will need to substitute the given values provided in the problem statement.
To solve this problem, we can use the conservation of momentum and the conservation of kinetic energy principles.
(a) First, let's find the velocity of the second block after the bullet embeds itself. According to the conservation of momentum:
Initial momentum = Final momentum
The initial momentum is the momentum of the bullet, and the final momentum is the sum of the momenta of the two blocks after the collision.
The momentum of the bullet can be calculated by multiplying its mass by its velocity:
Initial momentum = (mass of bullet) * (velocity of bullet)
Plugging in the values, we have:
Initial momentum = (7.45 g) * (+348 m/s)
Now, let's find the sum of the momenta of the blocks after the collision. The first block has a mass of 1220 g and a velocity of +0.648 m/s, while the second block has a mass of 1603 g and an unknown velocity (let's call it V2).
Final momentum = (mass of first block) * (velocity of first block) + (mass of second block) * (velocity of second block)
Plugging in the values, we have:
Final momentum = (1220 g) * (+0.648 m/s) + (1603 g) * (V2)
Since the bullet embeds itself in the second block, the two masses move together as one, so their velocities will have the same magnitude. Therefore, we can say that V2 is the velocity of the second block after the bullet embeds itself.
Now, equating the initial and final momenta, we can solve for V2:
(7.45 g) * (+348 m/s) = (1220 g) * (+0.648 m/s) + (1603 g) * (V2)
Now, we can solve for V2.
(b) To find the ratio of the total kinetic energy after the collision to that before the collision, we can use the conservation of kinetic energy. The kinetic energy before the collision is equal to the kinetic energy of the bullet, and the total kinetic energy after the collision is the sum of the kinetic energies of the two blocks.
The kinetic energy of an object can be calculated using the formula:
Kinetic energy = (1/2) * (mass) * (velocity)^2
Using this formula, we can calculate the kinetic energy before and after the collision and then find their ratio.