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 kinetic energy.

(a) Let's start by finding the velocity of the second block after the bullet embeds itself.

According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. The momentum is given by the product of mass and velocity.

Before the collision:
Momentum of the bullet = mass of the bullet * velocity of the bullet
Momentum of the first block = mass of the first block * velocity of the first block
Total momentum before the collision = Momentum of the bullet + Momentum of the first block

After the collision:
Momentum of the second block = mass of the second block * velocity of the second block

Using the conservation of momentum:

(mass of bullet * velocity of bullet) + (mass of first block * velocity of first block) = (mass of second block * velocity of second block)

Substituting the given values:

(7.45 g * 348 m/s) + (1220 g * 0.648 m/s) = 1603 g * velocity of second block

Solving for the velocity of the second block:

(7.45 g * 348 m/s) + (1220 g * 0.648 m/s) = 1603 g * velocity of second block
2595.6 g*m/s + 791.76 g*m/s = 1603 g * velocity of second block
3387.36 g*m/s = 1603 g * velocity of second block

Dividing both sides by 1603 g:

(3387.36 g*m/s) / 1603 g = velocity of second block
2.114 m/s = velocity of second block

Therefore, the velocity of the second block after the bullet embeds itself is +2.114 m/s (moving to the right).

(b) To find the ratio of the total kinetic energy after the collision to that before the collision, we need to calculate the kinetic energy in both cases.

Kinetic energy is given by the equation: KE = 1/2 * mass * velocity^2.

Before the collision, the total kinetic energy is the sum of the kinetic energies of the bullet and the first block:

KE_before = 1/2 * mass of bullet * (velocity of bullet)^2 + 1/2 * mass of first block * (velocity of first block)^2

Substituting the given values:

KE_before = 1/2 * 7.45 g * (348 m/s)^2 + 1/2 * 1220 g * (0.648 m/s)^2

Calculating the kinetic energy before the collision:

KE_before = 1/2 * 7.45 g * (121104 m^2/s^2) + 1/2 * 1220 g * (0.419904 m^2/s^2)
KE_before = 440045.2 g*m^2/s^2 + 325.9008 g*m^2/s^2
KE_before = 440371.1008 g*m^2/s^2

After the collision, the total kinetic energy is the kinetic energy of the second block:

KE_after = 1/2 * mass of second block * (velocity of second block)^2

Substituting the given values:

KE_after = 1/2 * 1603 g * (2.114 m/s)^2

Calculating the kinetic energy after the collision:

KE_after = 1/2 * 1603 g * (4.475896 m^2/s^2)
KE_after = 3586509.7896 g*m^2/s^2

The ratio of the total kinetic energy after the collision to that before the collision is:

KE_after / KE_before = (3586509.7896 g*m^2/s^2) / (440371.1008 g*m^2/s^2)

Calculating the ratio:

KE_after / KE_before ≈ 8.14

Therefore, the ratio of the total kinetic energy after the collision to that before the collision is approximately 8.14.

To solve this problem, we can use the principle of conservation of momentum and the principle of conservation of kinetic energy.

(a) To find the velocity of the second block after the bullet embeds itself, we need to use the conservation of momentum. The equation for conservation of momentum is:

m1 * v1i + m2 * v2i = m1 * v1f + m2 * v2f

Where:
m1 = mass of the bullet
v1i = initial velocity of the bullet
m2 = mass of the first block
v2i = initial velocity of the first block
v1f = final velocity of the bullet
v2f = final velocity of the second block

Plugging in the given values:

(7.45 g) * (+348 m/s) + (1220 g) * 0 = (7.45 g) * v1f + (1220 g + 1603 g) * v2f

Simplifying:

(7.45 g) * (+348 m/s) = (7.45 g) * v1f + (2823 g) * v2f

Now, solve for v2f:

v2f = [(7.45 g) * (+348 m/s) - (7.45 g) * v1f] / (2823 g)

Substituting the given value for v1f:

v2f = [(7.45 g) * (+348 m/s) - (7.45 g) * (0.648 m/s)] / (2823 g)

Calculating the value of v2f will give you the velocity of the second block after the bullet embeds itself.

(b) To find the ratio of the total kinetic energy after the collision to that before the collision, we use the conservation of kinetic energy principle. The equation for conservation of kinetic energy is:

(1/2) * m1 * v1i^2 + (1/2) * m2 * v2i^2 = (1/2) * m1 * v1f^2 + (1/2) * m2 * v2f^2

Where the variables have the same meaning as before.

Again, plugging in the given values:

(1/2) * (7.45 g) * (+348 m/s)^2 + (1/2) * (1220 g) * 0^2 = (1/2) * (7.45 g) * v1f^2 + (1/2) * (2823 g) * v2f^2

Simplifying:

(1/2) * (7.45 g) * (+348 m/s)^2 = (1/2) * (7.45 g) * v1f^2 + (1/2) * (2823 g) * v2f^2

Solving for the ratio of kinetic energies:

(total kinetic energy after) / (total kinetic energy before) = [(1/2) * (7.45 g) * v1f^2 + (1/2) * (2823 g) * v2f^2] / [(1/2) * (7.45 g) * (+348 m/s)^2]

Calculating this value will give you the ratio of the total kinetic energy after the collision to that before the collision.