A bullet of mass .03 kg moving with a speed of 400m/s penetrates 12cm into a fixed block of wood calculate the average force exerted by the wood on the bullet.
To calculate the average force exerted by the wood on the bullet, we need to use the principle of conservation of momentum. The bullet's initial momentum is mv, where m is the mass of the bullet and v is its initial velocity. Since the bullet penetrates into the wood and comes to rest, its final momentum is zero, meaning the change in momentum is -mv.
The average force exerted on an object is defined as the change in momentum divided by the time taken for the change to occur. However, in this case, we are not given the time taken for the bullet to come to rest inside the wood. Instead, we're given the distance the bullet penetrates, which we can use to find the stopping time.
To find the stopping time, we need to use the kinematic equation for
s = ut + (1/2)at^2,
where:
s = displacement (distance penetrated) = 12 cm = 0.12 m,
u = initial velocity = 400 m/s,
t = time taken,
a = acceleration = -u^2 / (2s), since the bullet comes to rest.
Plugging in the given values:
0.12 = (400)(t) + (1/2)(-400^2 / (2s)),
Simplifying:
0 = 400t - 800t,
1200t = 0.12,
t = 0.12 / 1200,
t = 0.001 s.
Now that we have the stopping time, we can find the average force using the formula:
Force = change in momentum / time taken.
The change in momentum is -mv, where m is the mass of the bullet and v is its initial velocity. In this case, m = 0.03 kg and v = 400 m/s.
Force = (-0.03 kg)(400 m/s) / 0.001 s.
Calculating this:
Force = -12,000 N.
Therefore, the average force exerted by the wood on the bullet is 12,000 Newtons.
Can you please explain it in simple way .
K.E. of bullet = Workdone
K.E. = 1/2 mv^2
Workdone =Force* Distance
1/2mv^2 = Force *0.12
1/2*0.03*(400)^2 = Force *0.12
Force = 20kn
V^2 = Vo^2 + 2a*d = 0.
400^2 + 2a*0.12 = 0,
160,000 + 0.24a = 0, a = -666,667 m/s^2.
F = M*a = 0.03 * (-666,667) = -20,000 N.