momentum is conserved
right is positive
(10 * 20) - (B * 20) = (B * 0) - (10 * 20)
2(10 * 20) = B * 20
How do I do this?
right is positive
(10 * 20) - (B * 20) = (B * 0) - (10 * 20)
2(10 * 20) = B * 20
The momentum of an object is given by the product of its mass and velocity. Therefore, the momentum of an object with mass m1 and velocity v1 is given by p1 = m1 * v1.
Let's denote the velocity of ball A after the collision as vA' and the mass of ball B as mB. We know the following information:
Ball A (before collision):
- Mass: mA = 10 kg
- Velocity: vA = 20 m/s to the right
Ball B (before collision):
- Velocity: vB = 20 m/s to the left
After collision:
- Ball B comes to a standstill (velocity is 0 m/s).
- Ball A moves to the left at twice its original speed: vA' = -40 m/s
Now, we can apply the law of conservation of momentum:
Total momentum before collision = Total momentum after collision
(mA * vA) + (mB * vB) = (mA * vA') + (mB * vB')
Substituting the given values:
(10 kg * 20 m/s) + (mB * (-20 m/s)) = (10 kg * (-40 m/s)) + (mB * 0 m/s)
Simplifying the equation:
200 kg路m/s - 20 mB = -400 kg路m/s
Now, let's solve for mB:
200 kg路m/s - 20 mB = -400 kg路m/s
Adding 20 mB to both sides:
200 kg路m/s = -400 kg路m/s + 20 mB
Simplifying further:
200 kg路m/s = -400 kg路m/s + 20 mB
Combining like terms:
600 kg路m/s = 20 mB
Dividing both sides by 20 m/s:
30 kg = mB
Therefore, the mass of ball B is 30 kg.