A 0.02kg bullet is fired at a 3.5kg block which is initially at resr on a table. The bullet embeds in the block. Friction converts the 24.02J of kinetic energy to heat and brings the block and bullet system to stop.
How fast was the bullet and block system sliding immediately after the collision?
i know its answer, which is 3.74, but i have no idea how does it can be 3.74
KE = 1/2 mv^2
Solve for v
I get 3.69
To find the speed of the bullet and block system immediately after the collision, we can use the principle of conservation of momentum.
First, we need to calculate the initial momentum of the bullet and block system. Momentum is calculated by multiplying the mass of an object by its velocity. Since the bullet is embedding in the block, we can assume that they move together as a single system after the collision.
Given:
Mass of the bullet (m1) = 0.02 kg
Mass of the block (m2) = 3.5 kg
Let's assume the velocity of the bullet and block system immediately after the collision as V.
The initial momentum (p) of the system before the collision can be calculated as:
p_initial = m1 * v1 + m2 * v2
where v1 is the initial velocity of the bullet and v2 is the initial velocity of the block (which is at rest, so v2 = 0).
Since the bullet and block system come to rest after the collision, the final momentum (p_final) is zero:
p_final = 0
According to the principle of conservation of momentum, the initial momentum should be equal to the final momentum:
p_initial = p_final
So, we can set up the equation:
m1 * v1 + m2 * v2 = 0
Substituting the known values, we get:
0.02 kg * v1 + 3.5 kg * 0 = 0
Simplifying the equation, we find:
0.02 kg * v1 = 0
Since the left side of the equation is equal to zero, we can conclude that the initial velocity of the bullet (v1) must also be zero.
Therefore, immediately after the collision, the bullet and block system is not sliding and has zero velocity.