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hodyreva [135]
3 years ago
8

A mass m = 75 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 19.2 m and finally

a flat straight section at the same height as the center of the loop (19.2 m off the ground). Since the mass would not make it around the loop if released from the height of the top of the loop (do you know why?) it must be released above the top of the loop-the-loop height. (Assume the mass never leaves the smooth track at any point on its path.)
Physics
1 answer:
dalvyx [7]3 years ago
4 0

Answer:

The velocity is 19.39 m/s

Solution:

As per the question:

Mass, m = 75 kg

Radius, R = 19.2 m

Now,

When the mass is at the top position in the loop, then the necessary centrifugal force is to keep the mass on the path is provided by the gravitational force acting downwards.

F_{C} = F_{G}

\frac{mv^{2}}{R} = mg

where

v = velocity

g = acceleration due to gravity

v = \sqrt{2gR} = \sqrt{2\times 9.8\times 19.2} = 19.39\ m/s

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When a golfer tees off, the head of her golf club, which has a mass of 160 g, is traveling 50 m/s just before it strikes a 46 g
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Answer:

v1=21.81m/s

Explanation:

<em>When a golfer tees off, the head of her golf club, which has a mass of 160 g, is traveling 50 m/s just before it strikes a 46 g golf ball at rest on a tee. Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 44 m/s. Neglect the mass of the club handle and determine the speed (in m/s) of the golf ball just after impact.</em>

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During collision of two particles, the external force on the system of two colliding particles is zero (only internal force acts between the colliding particles), therefore, the momentum is conserved during the collision.

Answer and Explanation:

Given :

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m1u1+m2u2=m1v1+m2v2.........................................1

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