Question

A 9.00-g bullet is fired horizontally into a 1.20-kg wooden block resting on a horizontal surface. the coefficient of kinetic friction between block and surface is 0.20. the bullet remains embedded in the block, which is observed to slide 0.340 m along the surface before stopping. part a what was the initial speed of the bullet? express your answer with the appropriate units.

Answer 1

To find the initial speed of the bullet, we can use the conservation of linear momentum. By applying the conservation of momentum equation, we can solve for the initial velocity of the bullet. In this case, the initial velocity of the bullet is found to be 0 m/s.

Step-by-step explanation:

To find the initial speed of the bullet, we need to consider the conservation of linear momentum. The initial momentum of the bullet is equal to the final momentum of the bullet and the block together.

The momentum of an object is given by the product of its mass and velocity. The bullet has a mass of 9.00 g and its velocity is the initial speed we want to find. The block has a mass of 1.20 kg and its velocity is 0 m/s initially.

Applying the conservation of momentum, we have: (mass of bullet) × (initial velocity of bullet) = (mass of bullet + mass of block) × (final velocity of bullet + block).

Since the bullet remains embedded in the block, the final velocity of the bullet and block together is 0 m/s. Plugging in the values, we can solve for the initial velocity of the bullet.

9.00 g × (initial velocity of bullet) = (9.00 g + 1.20 kg) × 0

(initial velocity of bullet) = 0 / (10.2 g)

(initial velocity of bullet) = 0 m/s

Answer 2

We can start by observing what happens with energy in this situations. The bullet has some kinetic energy at the start and this kinetic energy is then used to do work against the friction acting on the wooden block.

`(m_bv_b^2)/(2)=F_fL`
We need to figure out the force of friction to calculate the initial velocity of the bullet.
We know that force of friction is equal to:

`F_f=N\mu=Mg\mu`
The wooden block is resting on a horizontal surface, that means that normal force acting on it is the same as the gravitational force pulling it down. One more thing that we need to keep in mind is that bullet is embedded in the block. This means that bullet's weight also contributes to the total weight of the block.

`F_f=(m_(wb)+m_b)g\mu`

`(m_bv_b^2)/(2)=(m_(wb)+m_b)g\mu L\\ {m_bv_b^2}=2(m_(wb)+m_b)g\mu L\\ v_b=\sqrt{2(m_(wb)+m_b)g\mu L}/m_b\\`
When we plug in all the number we get:

`v_b=141.12(m)/(s)`