An explosion breaks a rock into three parts in a horizontal plane. Two...
An explosion breaks a rock into three parts in a horizontal plane. Two...
Since the rock breaks into three parts in a horizontal plane and two of them go off at right angles to each other, we can assume that the explosion imparts an initial velocity to each part in a perpendicular direction.
Let's denote the initial velocity of the first part (mass = 1 kg) as v1 = 12 m/s. Since the second part goes off at a right angle to the first part, its initial velocity would be perpendicular to the velocity of the first part. Let's denote the initial velocity of the second part (mass = m2) as v2.
Since there is no information given about the third part, we cannot determine its initial velocity.
Now, let's consider the conservation of momentum. In the absence of external forces, the total momentum before and after the explosion remains the same.
Before the explosion:
Total momentum = m1 * v1 + m2 * v2 + m3 * v3 (where m1, m2, m3 are the masses of the three parts, and v3 is the initial velocity of the third part)
After the explosion:
Total momentum = m1 * v1' + m2 * v2' + m3 * v3' (where v1', v2', v3' are the final velocities of the three parts)
Since the explosion breaks the rock into three parts, we can assume that it conserves mass. So, m1 + m2 + m3 = total mass of the rock.
Given that m1 = 1 kg and v1 = 12 m/s, we have:
Total momentum before the explosion = m1 * v1 + m2 * v2 + m3 * v3 = 1 kg * 12 m/s + m2 * v2 + m3 * v3
Now, if we assume that the explosion is the only force acting on the rock, then the total momentum after the explosion is zero (since there are no external forces to change the total momentum).
Therefore, Total momentum after the explosion = 0 = m1 * v1' + m2 * v2' + m3 * v3'
Since the second part moves at a right angle to the first part, we can assume that they have equal magnitudes of velocity. So, v1' = v2'.
Now, substituting the values and assuming v1' = v2' = v:
0 = 1 kg * v + m2 * v + m3 * v
0 = v(1 kg + m2 + m3)
Since the total mass of the rock is conserved, m1 + m2 + m3 = total mass of the rock, which we can denote as M.
Therefore, the equation becomes:
0 = v * M
Since the velocity v cannot be zero (as the parts move away from the explosion), the only way for the equation to be true is if M = 0.
However, it is not physically possible for the total mass of the rock to be zero. Therefore, the given scenario violates the conservation of momentum, and it is not a physically possible situation.