How can the kinetic energy ratio be calculated after a bullet collision?

• Physicsnoob90
In summary, a bullet with a mass of 6.64 g and a velocity of +340 m/s collides inelastically with a first block of mass 1234 g and velocity +0.577 m/s, passing through it and embedding itself in a second block of mass 1595 g. The velocity of the second block after the collision is 0.965 m/s. The ratio of the total kinetic energy after the collision to that before the collision is 2.48 x 10^-3.
Physicsnoob90

Homework Statement

A 6.64-g bullet is moving horizontally with a velocity of +340 m/s, where the sign + indicates that it is moving to the right. 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 1234 g, and its velocity is +0.577 m/s after the bullet passes through it. The mass of the second block is 1595 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.

The Attempt at a Solution

i've already found the answer for (a) Vf = 0.965 m/s, but I'm stuck in (b)

Steps:
1. KE(before) = 1/2(0.00664kg)(340 m/s)^2 = 3837.92
2. KE(after) = 1/2(1.60164kg)(0.965 m/s)^2 = 0.745743604

The Kinetic energy after should probably include the KE of the first moving block too.

So would the KE(after) = 1/2(1.234kg)(0.577m/s)^2 + 1/2(1.60164kg)(0.965 m/s)^2 = 0.951160797

Yeah, does that give you the right answer?

if i got my equation correct, the ratio should be:

KE(after)/KE(before) = 2.48x10^-4 (which i got wrong)

2.48 x 10^-3 if we include the key of the first block, 1.943 x 10^-3 if we don't.

Physicsnoob90
ah, you're right. i made a mistake on my sig figs. Thanks for the help!

1. What is bullet collision?

Bullet collision refers to the physical interaction between two objects, where one object (the bullet) strikes and potentially penetrates the other object.

2. How does bullet collision work?

When a bullet is fired from a gun, it travels at high speeds and carries a significant amount of kinetic energy. Upon impact with another object, this energy is transferred to the object, causing damage or penetration depending on the force and resistance of the object.

3. What factors influence bullet collision?

The speed, size, and shape of the bullet, as well as the material and thickness of the object it hits, are all important factors in determining the outcome of a bullet collision. Other factors such as angle of impact and environmental conditions can also play a role.

4. Can bullet collision be predicted?

While the outcome of a bullet collision cannot be predicted with 100% accuracy, it can be estimated using mathematical models and simulations based on the factors mentioned above. However, the unpredictable nature of real-world situations makes it difficult to accurately predict the exact outcome of a bullet collision.

5. How is bullet collision relevant in scientific research?

Bullet collision is relevant in a variety of fields, including ballistics, materials science, and forensic science. It can also be studied in laboratory settings to understand the effects of different variables on bullet penetration and damage, helping to improve safety measures and develop more effective materials and structures.

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