1) In any collision the momentum is conserved
(2*m)*(vo) + (m)*(-2*vo) = (2*m)(v1') + (m)(v2')
candel all the m factors (because they appear in all the terms on both sides of the equation)
2(vo) - 2(vo) = 2(v1') + (v2') => 2(v1') + v(2') = 0 => (v2') = - 2(v1')
2) Elastic collision => conservation of energy
=> [1/2] (2*m) (vo)^2 + [1/2](m)*(2*vo)^2 = [1/2](2*m)(v1')^2 + [1/2](m)(v2')^2
cancel all the 1/2 and m factors =>
2(vo)^2 + 4(vo)^2 = 2(v1')^2 + (v2')^2 =>
4(vo)^2 = 2(v1')^2 + (v2')^2
now replace (v2') = -2(v1')
=> 4(vo)^2 = 2(v1')^2 + [-2(v1')]^2 = 2(v1')^2 + 4(v1')^2 = 6(v1')^2 =>
(v1')^2 = [4/6] (vo)^2 =>
(v1')^2 = [2/3] (vo)^2 =>
(v1') = [√(2/3)]*(vo)
Answer: (v1') = [√(2/3)]*(vo)
Initial velocity=u=3m/s
Final velocity=v=10m/s
Time=t=8s
Using second equation of motion
Answer:
A. The force exerted by the sprinter must be 9.6 × 10² N.
B. The force that propels the sprinter is exerted by the blocks.
Explanation:
Hi there!
Let´s begin with part B:
The sprinter exerts a force on the blocks and, as a reaction, the blocks exert a force on the sprinter that is of equal magnitude but opposite direction (Newton´s third law). This reaction of the blocks causes the acceleration of the sprinter.
Part A
The force exerted by the blocks can be calculated using Newton´s second law:
F = m · a
Where:
F = exerted force.
m = mass of the object being accelerated.
a = acceleration of the object after applying the force on the object.
F = m · a
F = 64 kg · 15 m/s²
F = 9.6 × 10² N
The answer is True. The object is thrown away from the center of its original path due its inertia. Centrifugal force acts in the opposite direction as centripetal force. Centripetal force applies towards the center of the curvature in a spinning object. Centrifugal force is considered an apparent force while centripetal force is an actual force.
