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).
13 years ago
13 years ago
ps this is an inellastic question!
13 years ago
Goodness. That is why I gave you instructions to use conservation of Momentum only.
13 years ago
I got an answer of 3.16E7 m/s, is this right?
11 months ago
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