Answer:
7kgm/s
Explanation:
Using the law of conservation of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.
Let P1A and P1B be the initial momentum of the bodies A and B respectively
Let P2A and P2B be the final momentum of the bodies A and B respectively after collision.
Based on the law:
P1A+P2A = P1B + P2B
Given P1A = 5kgm/s
P2A = 0kgm/s(ball B at rest before collision)
P2A = -2.0kgm/s (negative because it moves in the negative x direction)
P2B = ?
Substituting the values in the equation gives;
5+0 = -2+P2B
5+2 = P2B
P2B = 7kgm/s
Answer:
4 s
Explanation:
Given:
Δx = 12 m
v₀ = 6 m/s
v = 0 m/s
Find: t
Δx = ½ (v + v₀) t
12 m = ½ (0 m/s + 6 m/s) t
t = 4 s
Answer:
, where the minus indicates the direction is opposite to that of the throw.
Explanation:
a)
Since MKS stands for meter-kilogram-second and we know that:
We can write that:
These are conversion factors, equal to 1, so multiplying our results by them won't change their value, only their units.
So we have that:
b)
Newton's 2nd Law tells us that F=ma, and the definition of acceleration is 
, so we have:
Taking the throw direction as the positive one, for our values we have:
