Could you please check my work?
2. A 2 kg ball of putty moving to the right at 3m/s has a head-on inelastic collision with a 1 kg ball of putty moving to the left at 3m/s. What is the final magnitude and direction of the velocity of the stuck together balls after the collision?
m1v1 +m2v2= (m1+m2)u2
(2kg)(3m/s) + (1kg)(3/s)= (2kg +1kg)u2
6kg/m/s +3kg/m/s= (3kg)u2
9kg/m/s= (3kg)u2
9kg/m/s divided by (3kg)= 3m/s and the direction would be to the left
No. the initial velocity v2 is -3m/s.
To check your work, let's calculate the final magnitude and direction of the velocity for the stuck together balls after the collision using the principles of conservation of momentum.
We start with the equation you used:
m1v1 + m2v2 = (m1 + m2)u2
Plugging in the given values, we have:
(2kg)(3m/s) + (1kg)(-3m/s) = (2kg + 1kg)u2
6kg m/s - 3kg m/s = 3kg u2
Now, let's simplify the equation:
3kg m/s = 3kg u2
Dividing both sides of the equation by 3kg, we find:
m/s = u2
So, the final magnitude of the velocity of the stuck together balls after the collision is 1 m/s. However, the direction should be to the right, not to the left. This is because the positive direction was set to the right in the problem statement.
Therefore, the correct answer is: The final magnitude of the velocity of the stuck together balls after the collision is 1 m/s to the right.