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Verdich [7]
4 years ago
5

Wil-E- Coyote drops a bowling ball off a cliff to try to catch the Roadrunner. The cliff is 132m high. What is its impact veloci

ty?​
Physics
1 answer:
omeli [17]4 years ago
3 0

Answer:

<u>Velocity impact is 50.86 m/s.</u>

Explanation:

If an "object of mass" "m" is dropped from "height" "h", then the "velocity" just before impact is  "v" . The "kinetic energy" before impact is equal to its "gravitational potential energy" at the height from which it was dropped: Kinetic energy = J.

\mathrm{v}^{2}=2 \mathrm{gh}

V is impact velocity

h height in meter = 132 m

\text { (gis referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^{2} \text { on Earth }

\mathrm{V}^{2}=2 \times 9.8 \times 132

V^{2}=19.6 \times 132

V^{2}=2587.2

V=\sqrt{2587.2}

V = 50.86 m/s

<u>Velocity impact is 50.86 m/s</u>.

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A 0.473 kg ice puck, moving east with a speed of 2.76 m/s, has a head-on collision with a 0.819 kg puck initially at rest. Assum
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The final speed of puck 1 is 0.739 m/s towards west  and puck 2 is 2.02 m/s towards east .

Explanation:

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mass of puck 2 , m_2= 0.819 kg

initial speed of puck 1 , u_1=2.76\frac{m}{s}

initial speed of puck 2 , u_2=0.00\frac{m}{s}

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Apply conservation of linear momentum

m_1u_1+m_2u_2=m_1v_1+m_2v_2

=>0.473\times 2.76+0.0=0.473\times v_1+0.819\times v_2

=>1.594=0.5775\times v_1+ v_2 -----(A)

Since collision is perfectly elastic , coefficient restitution e=1

u_2-u_1=v_1-v_2

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3 years ago
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