A particle of mass 5m moving with a speed v exploded and splits into two piece with masses of 2m and 3m. The lighter piece continue to move in the original dire
conservation of momentum
5 v = 2 V1 + 3 V2
Since you do not give the velocity of either after the collision, that is all I can say.
To solve this problem, let's start by applying the law of conservation of momentum:
The momentum before the explosion is equal to the momentum after the explosion.
Momentum before explosion = Momentum after explosion
Before explosion: The initial particle of mass 5m is moving with a speed v.
Momentum before explosion = (mass of particle 1) x (velocity of particle 1)
= (5m) x (v)
= 5mv
After explosion: The lighter piece has a mass of 2m and continues to move in the original direction. Let's call the velocity of the lighter piece v1.
The heavier piece has a mass of 3m. Since the total momentum is conserved, the velocity of the heavier piece can be calculated using the law of conservation of momentum.
Momentum after explosion = (mass of particle 1) x (velocity of particle 1) + (mass of particle 2) x (velocity of particle 2)
Using this equation, and substituting in the known values, we can solve for the velocity of the heavier piece:
5mv = (2m)(v1) + (3m)(velocity of the heavier piece)
Simplifying the equation:
5mv = 2mv1 + 3m(velocity of the heavier piece)
5v = 2v1 + 3(velocity of the heavier piece)
Since we want the lighter piece to move in the original direction, its velocity is the positive value v. Therefore, v1 = v.
5v = 2v + 3(velocity of the heavier piece)
5v - 2v = 3(velocity of the heavier piece)
3v = 3(velocity of the heavier piece)
velocity of the heavier piece = v
So, the velocity of the heavier piece after the explosion is equal to the velocity of the original particle, v.
In summary, the lighter piece with a mass of 2m continues to move in the original direction with a velocity of v, and the heavier piece with a mass of 3m also moves in the original direction with the same velocity, v.
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the explosion should be equal to the total momentum after the explosion.
Before the explosion, the initial momentum (p_initial) of the system is given by:
p_initial = m1 * v1 + m2 * v2
where m1 and m2 are the masses of the particle and v1 and v2 are their respective velocities.
Since the problem states that the original direction is maintained by the lighter piece, we can assume the velocities in the original direction to be v and 0 for the lighter and heavier pieces, respectively.
After the explosion, the final momentum (p_final) of the system is given by:
p_final = m1' * v1' + m2' * v2'
where m1' and m2' are the new masses of the two pieces, and v1' and v2' are their respective velocities after the explosion.
Using the principle of conservation of momentum, we can equate the initial and final momentum:
m1 * v + m2 * 0 = m1' * v1' + m2' * v2'
Since the lighter piece continues with the original velocity (v) and the heavier piece was at rest (0 velocity) before the explosion, we can simplify the equation to:
m * v = m1' * v + m2' * 0
Now, we can solve for m1' (mass of the lighter piece) by rearranging the equation:
m1' * v = m * v
m1' = m
Substituting the given values, the mass of the lighter piece (m1') is equal to 5m.
Therefore, the mass of the lighter piece after the explosion is 5m.