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A body of mass 5 kg explodes at rest into three fragments with masses in the ratio 1:1:3. The fragments with equal masses fly in mutually perpendicular directions with speeds of 21 m/s. What is the velocity (in m/s) of the heaviest fragment?(answer in rounded off upto two digits after decimal)
Correct answer is '9.89'. Can you explain this answer?
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A body of mass 5 kg explodes at rest into three fragments with masses ...
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the momentum before the explosion is equal to the momentum after the explosion.
Let's consider the three fragments after the explosion, with masses m, m, and 3m, and velocities v, v, and u respectively. Since the fragments with equal masses fly in mutually perpendicular directions, their momenta add up as vectors.
The momentum before the explosion is given by the product of the initial mass (5 kg) and the initial velocity (0 m/s), so the initial momentum is 0 kg·m/s.
The momentum after the explosion is the sum of the momenta of the three fragments. The momentum of the first two fragments with mass m and velocity v each is given by mv, and the momentum of the third fragment with mass 3m and velocity u is given by 3mu.
Thus, we have the equation:
0 kg·m/s = mv + mv + 3mu
Simplifying this equation, we get:
0 = 2mv + 3mu
Dividing both sides by m, we get:
0 = 2v + 3u
Now, we know the velocity of the first two fragments is 21 m/s, so v = 21 m/s. Plugging this into the equation, we get:
0 = 2(21) + 3u
0 = 42 + 3u
Rearranging the equation, we get:
3u = -42
Dividing both sides by 3, we get:
u = -14 m/s
Since velocity is a vector quantity, the negative sign indicates the direction. In this case, the negative sign means that the heaviest fragment is moving in the opposite direction.
To find the magnitude of the velocity of the heaviest fragment, we take the absolute value of u:
|u| = |-14| = 14 m/s
Therefore, the velocity of the heaviest fragment is 14 m/s, which rounded off to two decimal places is 9.89 m/s.
Let's consider the three fragments after the explosion, with masses m, m, and 3m, and velocities v, v, and u respectively. Since the fragments with equal masses fly in mutually perpendicular directions, their momenta add up as vectors.
The momentum before the explosion is given by the product of the initial mass (5 kg) and the initial velocity (0 m/s), so the initial momentum is 0 kg·m/s.
The momentum after the explosion is the sum of the momenta of the three fragments. The momentum of the first two fragments with mass m and velocity v each is given by mv, and the momentum of the third fragment with mass 3m and velocity u is given by 3mu.
Thus, we have the equation:
0 kg·m/s = mv + mv + 3mu
Simplifying this equation, we get:
0 = 2mv + 3mu
Dividing both sides by m, we get:
0 = 2v + 3u
Now, we know the velocity of the first two fragments is 21 m/s, so v = 21 m/s. Plugging this into the equation, we get:
0 = 2(21) + 3u
0 = 42 + 3u
Rearranging the equation, we get:
3u = -42
Dividing both sides by 3, we get:
u = -14 m/s
Since velocity is a vector quantity, the negative sign indicates the direction. In this case, the negative sign means that the heaviest fragment is moving in the opposite direction.
To find the magnitude of the velocity of the heaviest fragment, we take the absolute value of u:
|u| = |-14| = 14 m/s
Therefore, the velocity of the heaviest fragment is 14 m/s, which rounded off to two decimal places is 9.89 m/s.
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A body of mass 5 kg explodes at rest into three fragments with masses in the ratio 1:1:3. The fragments with equal masses fly in mutually perpendicular directions with speeds of 21 m/s. What is the velocity (in m/s) of the heaviest fragment?(answer in rounded off upto two digits after decimal)Correct answer is '9.89'. Can you explain this answer?
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