A bullet strike uniform plant with a velocity 400m/s snd comes out with half of that velocity. What should be the velocity of the out coming bullet.If the plant were half thick?
If the plant were half thick, the bullet would probably think twice before striking it! But if it still mustered the courage to strike, I guess it would come out with the same velocity it went in with, since thickness doesn't really affect the speed of a bullet. It's always a good reminder that plants are better off growing peacefully without any disturbance from bullets!
To find the velocity of the bullet as it comes out if the plant were half as thick, we can apply the principle of conservation of momentum.
According to the principle of conservation of momentum, the total momentum before an event should be equal to the total momentum after the event, assuming no external forces acting on the system.
Let's denote the original velocity of the bullet as V1 (400 m/s) and the final velocity as V2. Since the bullet comes out with half the velocity, we can express it as V1/2.
Now, let's assume the mass of the bullet is M, and the mass of the plant is P. If the plant were half as thick, the mass of the plant would also be half, so we can consider it as P/2.
According to the conservation of momentum, the total momentum before the event is equal to the total momentum after the event. In this case, we can write the equation as:
M * V1 = M * (V1/2) + (P/2) * V2
Simplifying the equation, we get:
V2 = (2 * M * V1 - M * (V1/2)) / (P/2)
Now, in this scenario, we don't have the exact values for the mass of the bullet (M) and the mass of the plant (P). So, you need to provide the values for those variables, and we can calculate the velocity of the outcoming bullet using the equation mentioned above.
To find the velocity of the outgoing bullet, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision should be equal to the total momentum after the collision.
Let's denote the initial velocity of the bullet as v1 and the final velocity of the bullet as v2. Initially, the bullet is moving with a velocity of 400 m/s, so v1 = 400 m/s.
According to the question, the bullet strikes a uniform plant and comes out with half of its velocity. This means that the final velocity of the bullet (v2) is half of the initial velocity (v1), so v2 = v1/2.
Now, let's consider the conservation of momentum. The total momentum before the collision is given by the mass of the bullet (m) multiplied by its initial velocity (v1). The total momentum after the collision is given by the mass of the bullet multiplied by its final velocity (v2).
Therefore, we can write the equation:
m * v1 = m * v2
Since the mass of the bullet is the same before and after the collision, we can cancel it out from both sides of the equation:
v1 = v2
Substituting the values of v1 and v2:
400 m/s = (400 m/s) / 2
Now we can solve for v2:
v2 = 400 m/s / 2
v2 = 200 m/s
Therefore, the velocity of the outgoing bullet, if the plant were half thick, would be 200 m/s.
thickness=s
acceleration of the bullet inside plank=-a
200^2-400^2=-2as
as=60000 units
v^2-400^2=2*a*s/2
v^2=160000-60000
=100000 units
v=100root10 m/s