Question

A 3.00 g bullet is fired horizontally into a 2.00 kg wooden block resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.30. The bullet remains embedded in the block, which is observed to slide 0.20 m along the surface before stopping. What was the initial speed of the bullet?

Answer 1

• Mass of bullet, m=6g=0.006 Kg

• Mass of gun M= 1.5 Kg

• Distance covered by the block d= 0.25m.

If u is the speed of bullet and v is the the final speed of bullet and block system then:

• mu+0=(m+M)v=0.006u=1.506v ; v=0.00398u

Work done by friction = change in kinetic energy of block:

• fd= 0-1/2mv^2=
• 0-1/2mv^2= umg
• u=608m/s

Step-by-step explanation:

Answer 2

To determine the initial speed of the bullet, the work-energy principle and momentum conservation are used after calculating the work done by friction to bring the block-bullet system to rest.

Step-by-step explanation:

The student is asking about the initial speed of a bullet that fired into a wooden block. This is a classic physics problem that involves concepts such as momentum conservation and work-energy principles. To find the initial speed of the bullet, the following steps are used:

• Firstly, use the work-energy principle to calculate the kinetic energy of the block-bullet system just after the collision since it is equal to the work done by friction to stop the block.
• The work done by friction (which equals the kinetic energy) can be found using the formula W = fd where W is work, f is frictional force, and d is the distance slide.
• Frictional force can be calculated from f = μkn where μk is the coefficient of kinetic friction and n is the normal force, which equals the weight of the block for horizontal surfaces.
• Using momentum conservation, the initial momentum of the bullet equals the final momentum of the bullet-block system, and from this, the bullet's initial speed can be derived.