A frisbee is flying through the air. Which forces are acting on it (choose 3)
1 answer:
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Answer:
(a) 2.03 m/s
(b) 2.15 m/s
Explanation:
(a)
From the law of conservation of momentum, the sum of initial momentum equals the sum of final momentum
Momentum, p=mv where m is the mass and v is the velocity
where
is the common velocity,
and
are velocities of the first railroad car and the second railroad car respectively,
and
are masses of the first railroad car and the second railroad car respectively
Substituting the given figures then
(b)
Momentum, p=mv where m is the mass and v is the velocity
where
and
are initial velocities of the ball moving towards west and the other ball moving in opposite direction respectively,
and
are masses of the first railroad car and the second railroad car respectively
Taking west as positive then the opposite direction will be negative hence
Note that when the 0.455 rebounds, it moves towards East and since we took West as positive that is why we give 15.4 as a negative value.
towards East
Answer:
6.86 m/s
Explanation:
This problem can be solved by doing the total energy balance, i.e:
initial (KE + PE) = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}
Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.
Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.
Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s
The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²
This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)
Now, 1/6m(v0)² = mgL ⇒ v0 =
Hence, v0 = 6.86 m/s