Question

A bullet of mass m is fired at speed v0 into a wooden block of mass M. The bullet instantaneously comes to rest in the block. The block with the embedded bullet slides along a horizontal surface with a coefficient of kinetic friction μ. Write an expression for how far the block slides (s) in terms of m, M, Vo , μ, and g.

Answer 1

The distance the block slides (s) after a bullet embeds in it is given by the expression s = 1/2(m·v0)² / (μ·g·(M + m)), derived by using conservation of momentum and the work-energy principle.

Step-by-step explanation:

To solve how far the wooden block slides (s) after a bullet is fired into it, we'll use the conservation of momentum and work-energy principles. First, we determine the velocity of the block immediately after the collision using the conservation of momentum:

m·v0 = (M + m)·v

Where m is the mass of the bullet, v0 is the initial velocity of the bullet, M is the mass of the block, and v is the final velocity of the block and bullet system. After the collision, the kinetic energy of the block and bullet system will be dissipated by the work done against friction:

½(M + m)·v2 = μ(M + m)g·s

We can then solve for s giving us:

s = ½v2 / (μ·g)

Substitute v from the first equation into the second yields:

s = ½(m·v0)2 / (μ·g·(M + m))

This is the expression for the distance s the block slides in terms of the given variables.

Answer 2

`s = (1)/(2\mu g)\left ((mv_(0))/(m + M)  \right )^(2)`

Step-by-step explanation:

mass of bullet = m

mass of block = M

initial speed of bullet = vo

coefficient of friction = μ

Let the horizontal distance traveled before stopping is s.

Let v be the velocity of the bullet and block system after collision.

Use conservation of momentum

m x vo = (M+m) x v

`v = (mv_(0))/(m + M)`

Now use third equation of motion

v² = u² - 2 a s

So,

`0 = \left ((mv_(0))/(m + M)  \right )^(2)-2\mu gs`

`s = (1)/(2\mu g)\left ((mv_(0))/(m + M)  \right )^(2)`