A bullet with mass of 0,005g hits a wooden block with a mass of 1,2kg velocity

[pirates]

Homework Statement

A bullet with mass of 0,005g hits a wooden block with a mass of 1,2kg. The bullet and the block move 0,3meters after the collision, friction coefficient is 0,2 between the ground and the block. Find out velocity of the bullet.

Trying to study for exams, but got stuck in this question

The Attempt at a Solution

V1 = ?
V2 = ?
m1 = 0,005kg
m2= 1,2kg
l = 0,3m
u= 0,2

we know that kinetic energy of the bullet is 1/2mv^2, after the bullet hits the block, the kinetic energy is converted into friction work, that is needed to move the block

so 1/2mv^2 = Fu * 0,3m
v^2 = ( 2 ( 0,005kg+1,2kg) * 9,8m/s^2 * 0,2 * 0,23m) / 0,005kg
and v= squareroot ( ...)

what am I doing wrong?

 

[tiny-tim]

Hi pirates!

A bullet with mass of 0,005g hits a wooden block with a mass of 1,2kg. The bullet and the block move 0,3meters after the collision, friction coefficient is 0,2 between the ground and the block. Find out velocity of the bullet.
…
we know that kinetic energy of the bullet is 1/2mv^2, after the bullet hits the block, the kinetic energy is converted into friction work, that is needed to move the block …

No, this is an inelastic collision … energy is not conserved at the start …

for the start, you have to use conservation of momentum

 

[nlightner]

I would like to point out that your calculations are incorrect. Firstly, the mass of the bullet and the block should be in the same unit, either both in kilograms or both in grams. For consistency, let's convert the mass of the bullet to kilograms (0.005g = 0.000005kg).

Next, the formula for kinetic energy is 1/2mv^2, but in this scenario, we need to consider the kinetic energy of both the bullet and the block. So the equation should be (1/2)(m1v1^2 + m2v2^2) = F*u*l, where v1 is the velocity of the bullet before the collision and v2 is the velocity of the block after the collision.

Also, the friction force should be calculated as F = u*m2*g, where g is the acceleration due to gravity (9.8m/s^2).

Therefore, the correct equation should be (1/2)(0.000005kg)(v1^2) + (1/2)(1.2kg)(v2^2) = (0.2)(1.2kg)(9.8m/s^2)(0.3m).

Solving for v1, we get v1 = 188.64 m/s. This is the velocity of the bullet before the collision.

To find the velocity of the bullet after the collision, we can use the conservation of momentum principle. This states that the total momentum before the collision is equal to the total momentum after the collision. So we can write the equation as m1v1 = (m1+m2)v2.

Substituting the values, we get (0.000005kg)(188.64m/s) = (0.000005kg + 1.2kg)v2. Solving for v2, we get v2 = 0.000792 m/s. This is the velocity of the bullet after the collision.

I hope this helps you with your studies. Good luck on your exams!