We can do this with the conservation of momentum. The fact it is elastic means no KE is lost so we don't have to worry about the loss due to sound energy etc.
Firstly, let's calculate the momentum of both objects using p=mv:
Object 1:
p = 0.75 x 8.5 = 6.375 kgm/s
Object 2 (we will make this one negative as it is travelling in the opposite direction):
p = 0.65 x -(7.2) = -4.68 kgm/s
Based on this we know that the momentum is going to be in the direction of object one, and will be 6.375-4.68=1.695 kgm/s
Substituting this into p=mv again:
1.695 = (0.75+0.65) x v
Note I assume here the objects stick together, it doesn't specify - it should!
1.695 = 1.4v
v=1.695/1.4 = 1.2 m/s to the right (to 2sf)
Let F = required force, N
Given:
d = 12 m, distance
W = 280 J, work done
By definition,
W = F*d,
therefore
(F N)*(12 m) = (280 J)
F = 280/12 = 23.33 N
Answer: The force is 23.3 N (nearest tenth)
_ acceleration occurs when an object speeds up.
Answer
Positive
It’s my guess but from my opinion i would say yes
Answer:
12N
Explanation:
Suppose the string mass is negligible, the total mass of the 2 block system is 6 + 9 = 15 kg
So the acceleration of the system when subjected to 30N force is
a = F / M = 30 / 15 = 2 m/s2
So both blocks would have the same acceleration, however, the force acting on the 6kg block would have a magnitude of
f = am = 2 * 6 = 12N
This is the tension in the string between the blocks
