Car crosses the first third of the road with unknown speed and the rest of the road with the speed of 50 km/h. What is the the s
1 answer:
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Answer:
w = vR/3
Explanation:
The centre of mass of the loop to bullet system is given by D / 4 from centre of loop, which is equivalent to R / 2 from its centre.
From the principle of conservation of linear momentum , we have
m*v = 2*m* Vcm
Where v = velocity of bullet, Vcm = velocity of wood
Hence, we have
Vcm = v2
Also, from the conservation of angular momentum about the centre of mass.
M*V*(R/2) = Ic*w - equation (I)
where Ic = moment of inertia and w = angular velocity
Ic for a ring is given by
Ic of a bullet is given by
Hence, the moment of inertia of the system is given by the summation of the two moments of inertia Ic(ring) + Ic(bullet) which gives
Ic(system) =
Substituting back into equation (I), we have
Hence, we obtain w =vR/3
w=v3R
Answer:
Approximately (assuming that
, the ball was launched from ground level, and that the drag on the ball is negligible.)
Explanation:
Let denote the velocity at which the ball was thrown upward.
If the drag (air friction) on the ball is negligible, the ball would land with a velocity of exactly . The velocity of the ball would be changed from
to
(such that
) within
.
Also because the drag on the ball is negligible, the acceleration of the ball would be . Thus:
.
Since :
.
.
The ball reaches maximum height when its velocity is . Apply the SUVAT equation
to find the displacement
between the original position (ground level, where
) and the max-height position of the ball (where
.)
.