A particle with m = 3.3E-27 kg is moving with a velocity of 6.0E7m/s. It then collides with a stationary particle of mass 2m, the lighter particle then moves at a right angle to the original direction with a velocity of 2.0E7m/s. What is the veloctiy of the heavier particle with (mass 2m).
The heaviercomponent has the original momentum of the moving partical in the horizontal directoni, and the heavier particlehas the opposite momentum vertically to the original particle going up.
ps this is an inellastic question!
Goodness. That is why I gave you instructions to use conservation of Momentum only.
I got an answer of 3.16E7 m/s, is this right?
To solve this problem, we can use the principle of conservation of momentum, which states that the total momentum before a collision is equal to the total momentum after the collision.
1. First, let's calculate the initial momentum of the system before the collision.
Momentum (p) is given by the formula: p = m * v, where m is the mass and v is the velocity.
For the lighter particle: p1 = (3.3E-27 kg) * (6.0E7 m/s)
For the heavier particle: p2 = (2 * 3.3E-27 kg) * 0 (since it is stationary)
Therefore, the total initial momentum (p_initial) is p1 + p2.
2. Next, let's calculate the final momentum of the system after the collision.
After the collision, the lighter particle moves at a right angle to the original direction, while the heavier particle acquires a velocity (v2).
For the lighter particle: p1f = (3.3E-27 kg) * (2.0E7 m/s) direction (perpendicular to the original direction)
For the heavier particle: p2f = (2 * 3.3E-27 kg) * (v2) direction (v2 is the velocity we need to find)
Therefore, the total final momentum (p_final) is p1f + p2f.
3. According to the principle of conservation of momentum, p_initial = p_final.
Equating the initial and final momentum, we get:
p1 + p2 = p1f + p2f
(3.3E-27 kg) * (6.0E7 m/s) + (2 * 3.3E-27 kg) * 0 = (3.3E-27 kg) * (2.0E7 m/s) + (2 * 3.3E-27 kg) * (v2)
4. Solve the equation for the velocity (v2) of the heavier particle:
(3.3E-27 kg) * (6.0E7 m/s) = (3.3E-27 kg) * (2.0E7 m/s) + (2 * 3.3E-27 kg) * (v2)
3.3E-27 kg * 6.0E7 m/s = 3.3E-27 kg * 2.0E7 m/s + 2 * 3.3E-27 kg * v2
Simplifying the equation gives: 3.3E-27 kg * v2 = (3.3E-27 kg * 6.0E7 m/s) - (3.3E-27 kg * 2.0E7 m/s)
v2 = [(3.3E-27 kg * 6.0E7 m/s) - (3.3E-27 kg * 2.0E7 m/s)] / (3.3E-27 kg * 2)
v2 = (3.3E-27 kg * 4.0E7 m/s) / (3.3E-27 kg * 2)
v2 = 2.0E7 m/s
So, the velocity of the heavier particle (mass 2m) is 2.0E7 m/s