Collisions - bullet shot in in block

In summary, three blocks of different masses are arranged linearly on a frictionless surface. A bullet is shot at the first block and travels through it, embedding itself into the second block which then collides into the third block. The questions asked are: a) the velocity of block 2 just before it collides into block 3 in an elastic final collision, b) the initial velocity of the bullet before it embeds into block 2, c) the speed of block 1 after the bullet travels through it if it slowed down by 10%, and d) the percentage of initial energy lost by the bullet. The formula used for the calculations is (2m1/m2+m2)* v1, but the resulting numbers
  • #1
Ajmathusek
3
0
Three blocks (1,2,3) of mass 3.00 kg, 2.00 kg, 1.00 kg sit
linearly 10.0 cm apart from each other on a frictionless surface.
A 25 g bullet is shot at the first block, it travels through the first
block, and embeds itself into the second block (without falling over) which then collides into the
third block.
a. Assuming the final collision is elastic, and the third block has a final velocity of 5.00 m/s, what is
the velocity of block 2 just before it collides into block 3?
b. What is the velocity of the bullet before it embeds itself into block 2?
c. Assuming the bullet slowed down by only 10% while it traveled through block 1, what was the
speed of block 1 after the bullet traveled through it?
d. What percent of the initial energy of the bullet was lost?

I tried using the formula (2m1/m2+m2)* v1 for all the question in the problem. But the numbers don't sound right to me.
 
Last edited:
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  • #2
Ajmathusek said:
I tried using the formula (2m1/m2+m2)* v1 for all the question in the problem. But the numbers don't sound right to me.
Please show your calculations and numbers then. Otherwise we have no way to tell what went wrong. Your formula looks a bit odd, but that could be a typo.

Note that some collisions are not elastic.
 

What is a collision?

A collision is an event in which two or more objects come into contact with each other and exchange energy. In the case of a bullet shot in a block, the bullet and block will collide and the bullet's kinetic energy will be transferred to the block, causing it to move.

What factors affect the outcome of a collision between a bullet and a block?

The outcome of a collision between a bullet and a block can be affected by factors such as the mass of the bullet and block, the velocity of the bullet, the type of material the block is made of, and the angle at which the bullet hits the block.

How does the conservation of momentum apply to a collision between a bullet and a block?

The conservation of momentum states that the total momentum of a system remains constant before and after a collision. In the case of a bullet shot in a block, the total momentum of the bullet and block before the collision will be equal to the total momentum after the collision.

Can the energy lost in a collision between a bullet and a block be calculated?

Yes, the energy lost in a collision between a bullet and a block can be calculated using the principle of conservation of energy. The initial kinetic energy of the bullet will be equal to the final kinetic energy of the block after the collision, taking into account any energy lost due to friction or deformation of the block.

What are some real-life applications of studying collisions between a bullet and a block?

Studying collisions between a bullet and a block can have real-life applications in fields such as ballistics and forensic science. It can also help engineers design more efficient and safe structures to withstand high-velocity impacts, such as bulletproof vests and armored vehicles.

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