a bomb of mass 5 kg explodes in to two equel fragments. At what angle these fragment fly apart ?
180 degrees because momentum is conserved and there is no outside force their velocities must sum to zero.
To determine the angle at which the two fragments fly apart after a bomb explosion, we need to consider the conservation of momentum. In an explosion or any other isolated system, the total momentum before the event should be equal to the total momentum after the event.
Let's assume that the initial velocity of the bomb is zero, and after the explosion, it breaks into two fragments of equal mass, m, moving at angles θ1 and θ2 respectively with respect to the original line of motion.
Using the conservation of momentum, we can write the equation:
Initial momentum = Final momentum
0 = m * v1 + m * v2
Since the fragments have equal mass and the initial velocity is zero, the equation simplifies to:
0 = v1 + v2
Now, let's consider the momentum components in the x-direction and y-direction.
In the x-direction, the total momentum before and after the explosion should be zero since there is no initial velocity, and the fragments are moving along different angles. Therefore:
0 = m * v1 * cos(θ1) + m * v2 * cos(θ2)
Similarly, in the y-direction:
0 = m * v1 * sin(θ1) + m * v2 * sin(θ2)
Now, we can use these two equations to solve for the angles θ1 and θ2.
Simplifying the x-direction equation:
0 = v1 * cos(θ1) + v2 * cos(θ2)
This equation indicates that the sum of the x-components of the velocities should be zero.
Simplifying the y-direction equation:
0 = v1 * sin(θ1) + v2 * sin(θ2)
This equation indicates that the sum of the y-components of the velocities should be zero.
By solving these two equations simultaneously, we can find the values of θ1 and θ2, which represent the angles at which the fragments fly apart. Keep in mind that there may be multiple solutions depending on the specific values of the velocities.