A 20 kg mass is moving at 10 m/s and collides with a stationary 5 kg mass, transferring all its momentum in the collision, what
1 answer:
Answer:
v = 40 [m/s].
Explanation:
Linear momentum is defined as the product of mass by Velocity. In this way, by means of the following equation, we can calculate the momentum.
where:
m = mass [kg]
v = velocity [m/s]
Since all momentum is transferred, we can say that this momentum is equal for the mass of 5 [kg]. In this way, we can determine the speed after the impact.
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Answer:
Fc = [ - 4.45 * 10^-8 j ] N
Explanation:
Given:-
- The masses and the position coordinates from ( 0 , 0 ) are:
Sphere A : ma = 80 kg , ( 0 , 0 )
Sphere B : ma = 60 kg , ( 0.25 , 0 )
Sphere C : ma = 0.2 kg , ra = 0.2 m , rb = 0.15
- The gravitational constant G = 6.674×10−11 m3⋅kg−1⋅s−2
Find:-
what is the gravitational force on C due to A and B?
Solution:-
- The gravitational force between spheres is given by:
F = G*m1*m2 / r^2
Where, r : The distance between two bodies (sphere).
- The vector (rac and rbc) denote the position of sphere C from spheres A and B:-
Determine the angle (α) between vectors rac and rab using cosine rule:
Determine the angle (β) between vectors rbc and rab using cosine rule:
- Now determine the scalar gravitational forces due to sphere A and B on C:
Between sphere A and C:
Fac = G*ma*mc / rac^2
Fac = (6.674×10−11)*80*0.2 / 0.2^2
Fac = 2.67*10^-8 N
vector Fac = Fac* [ - cos (α) i + - sin (α) j ]
vector Fac = 2.67*10^-8* [ - cos (36.87°) i + -sin (36.87°) j ]
vector Fac = [ - 2.136 i - 1.602 j ]*10^-8 N
Between sphere B and C:
Fbc = G*mb*mc / rbc^2
Fbc = (6.674×10−11)*60*0.2 / 0.15^2
Fbc = 3.56*10^-8 N
vector Fbc = Fbc* [ cos (β) i - sin (β) j ]
vector Fbc = 3.56*10^-8* [ cos (53.13°) i - sin (53.13°) j ]
vector Fbc = [ 2.136 i - 2.848 j ]*10^-8 N
- The Net gravitational force can now be determined from vector additon of Fac and Fbc:
Fc = vector Fac + vector Fbc
Fc = [ - 2.136 i - 1.602 j ]*10^-8 + [ 2.136 i - 2.848 j ]*10^-8
Fc = [ - 4.45 * 10^-8 j ] N
Explanation:
It is given that,
Mass of person, m = 70 kg
Radius of merry go round, r = 2.9 m
The moment of inertia,
Initial angular velocity of the platform,
Part A,
Let is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :
Here,
Solving the above equation, we get the value as :
Part B,
The initial rotational kinetic energy is given by :
The final rotational kinetic energy is given by :
Hence, this is the required solution.