a particle of mass 5m moving with speed v, explodes and splits into two pieces with masses of 2m and 3m. The lighter piece continues to move in the original direction with speed v relative to the heavier piece. What is the speed of the lighter piece
The total momentum remains 5m*v in the forward direction.
Heavier piece speed after breakup = v1
Lighter piece speed after breakup = v2
v2 - v1 = v
5mv = 2m*v2 + 3m*v1
= 2m*v2 + 3m*(v2 - v)
8*mv = 5*mv2
v2 = (8/5)v
v1 = (3/5)v
The lighter piece moves faster.
To find the speed of the lighter piece after the explosion, we can apply 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.
The momentum of an object is defined as the product of its mass and velocity: momentum = mass * velocity.
Before the explosion, the momentum of the initial particle (with mass 5m and velocity v) can be calculated as: momentum_before = (5m) * v.
After the explosion, the momentum of the lighter piece (with mass 2m) can be calculated as: momentum_lighter = (2m) * v_lighter
Similarly, the momentum of the heavier piece (with mass 3m) can be calculated as: momentum_heavier = (3m) * v_heavier
According to the conservation of momentum, the sum of the momenta before the explosion should be equal to the sum of the momenta after the explosion: momentum_before = momentum_lighter + momentum_heavier
Now we can substitute the respective momenta into the equation: (5m) * v = (2m) * v_lighter + (3m) * v_heavier
Since the lighter piece continues to move in the original direction with speed v relative to the heavier piece, we can express the velocity of the lighter piece as: v_lighter = v_heavier + v
Substituting this expression into the equation, we get: (5m) * v = (2m) * (v_heavier + v) + (3m) * v_heavier
Simplifying this equation gives: 5mv = 2mv_heavier + 2mv + 3mv_heavier
Combining like terms, we have: 5mv = 5mv_heavier + 2mv
Simplifying, we get: 0 = 5mv_heavier - 3mv
Rearranging the equation gives: 3mv = 5mv_heavier
Dividing both sides by 3m, we get: v = (5/3) * v_heavier
Therefore, the speed of the lighter piece (v_lighter) is equal to (5/3) times the speed of the heavier piece (v_heavier).
To find the speed of the lighter piece after the explosion, we can use the principle of conservation of linear momentum. According to this principle, the total momentum of an isolated system remains constant before and after an explosion.
Before the explosion, the momentum of the system is given by:
Momentum before = mass of particle (5m) × velocity of particle (v)
After the explosion, the total momentum of the system is still the same, but it is now divided between the two pieces. Let's denote the velocity of the lighter piece as v1 and the velocity of the heavier piece as v2.
The momentum of the lighter piece (2m) will be:
Momentum of lighter piece = mass of lighter piece (2m) × velocity of lighter piece (v1)
The momentum of the heavier piece (3m) will be:
Momentum of heavier piece = mass of heavier piece (3m) × velocity of heavier piece (v2)
Since the lighter piece continues to move in the original direction with speed v relative to the heavier piece, we have:
v1 = v + v2
Using the conservation of momentum equation, we can write:
Momentum before = Momentum of lighter piece + Momentum of heavier piece
Therefore,
mass of particle (5m) × velocity of particle (v) = mass of lighter piece (2m) × velocity of lighter piece (v1) + mass of heavier piece (3m) × velocity of heavier piece (v2)
Substituting v1 = v + v2 into this equation, we get:
5m × v = 2m × (v + v2) + 3m × v2
Simplifying this equation, we have:
5mv = 2mv + 2mv2 + 3mv2
Combining like terms, we get:
0 = 5mv + 5mv2
Dividing both sides by 5m, we obtain:
0 = v + v2
Since we are looking for the speed of the lighter piece (v1), which is v + v2, we can conclude that the speed of the lighter piece is 0.
Therefore, after the explosion, the lighter piece comes to a stop and has a speed of 0.