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 0.310 m along the surface before stopping. (a) What was teh initial speed of the bullet?
(b) What kind of collision took place between the bullet and the block? Explain your answer comparing the values of the kinetic energy before and after the collision.
(c) Calculate the impulse of the block from the moment just after the impact to his final position. Analyze the result.

Answer 1

a) The initial speed of the bullet is 330.5 m/s

b) The collision is inelastic

c) The impulse is -2.97 kg m/s

Explanation:

a)

The energy lost by the block while sliding (which is equal to the work done by friction) is equal to the kinetic energy of the block after the bullet has been embedded into it, therefore we can write

$KE=W$

$\frac{1}{2}(M+m)v^2=(\mu (M+m)g) d$

where

M = 1.20 kg is the mass of the block

m = 9.00 g = 0.009 kg is the mass of the bullet

v is the combined speed of bullet+block after the collision

$\mu = 0.20$ is the coefficient of friction

$g=9.8 m/s^2$ is the acceleration of gravity

d = 0.310 m is the distance through which the block slides

Solving for v,

$v=\sqrt{2gd}=\sqrt{2(9.8)(0.310)}=2.46 m/s$

This is the final velocity of the block+bullet after the collision.

Now we can apply the law of conservation of momentum: in fact, the total momentum of the system before the collision must be equal to the total momentum after the collision, so we get

$mu+MU = (m+M)v$

where

u is the initial velocity of the bullet

U = 0 is the initial velocity of the block (initially at rest)

v = 2.46 m/s

Solving for u,

$u=\frac{(m+M)v}{m}=\frac{(0.009+1.20)(2.46)}{0.009}=330.5 m/s$

b)

To check whether the collision is elastic or inelastic, we just need to compare the total kinetic energy before and after the collision.

Before the collision, we have:

$K_i = \frac{1}{2}mu^2 = \frac{1}{2}(0.009)(330.5)^2=491.5 J$

While after the collision

$K_f = \frac{1}{2}(m+M)v^2 = \frac{1}{2}(0.009+1.20)(2.46)^2=3.7 J$

We see that the final kinetic energy is less than the initial kinetic energy: therefore, the collision is inelastic, since part of the energy has been converted into other forms of energy (e.g. thermal energy).

c)

The impulse of the block is equal to its change in momentum, so:

$I=\Delta p =p_f - p_i = (m+M)v'-(m+M)v$

where

v' = 0, since the block comes to a stop

v = 2.46 m/s is the velocity of the block just after the collision

Substituting,

$I=0-(0.009+1.20)(2.46)=-2.97 kg m/s$

And the impulse is negative, because its direction is opposite to the direction of motion of the block (this means that the force exerted on the block, which is the force of friction, acts in the direction opposite to the motion of the block).

Learn more about kinetic energy and momentum:

brainly.com/question/6536722

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly