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grin007 [14]
3 years ago
14

One kind of slingshot consists of a pocket that holds a pebble and is whirled on a circle of radius r. The pebble is released fr

om the circle at the angle θ so that it will hit the target. The angle ϕ in the drawing is 37.1°. The distance to the target from the center of the circle is d. (See the drawing below, which is not to scale.) The circular path is parallel to the ground, and the target lies in the plane of the circle. The distance d is one times the radius r. Ignore the effect of gravity in pulling the stone downward after it is released and find the angle θ.

Physics
1 answer:
Debora [2.8K]3 years ago
5 0

Answer:

θ=142.9°

Explanation:

d=1 *r

angle ϕ= 37.1°

the line connecting pebble and target should be tangent to a circle so

cos(180-ϕ-θ)=\frac{r}{d}=\frac{1}{1}

∴ θ=180-ϕ-cos^{-1} (\frac{1}{1} )

  θ= 180-37.1-0

  θ=142.9°

 

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On a cold winter day, a penny (mass 2.50 g) and a nickel (mass 5.00 g) are lying on the smooth (frictionless) surface of a froze
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0.78333 m/s in the opposite direction

1.566 m/s in the same direction

Explanation:

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m_2 = Mass of nickel = 0.005 kg

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u_2 = Initial Velocity of nickel = 0 m/s

v_1 = Final Velocity of penny

v_2 = Final Velocity of nickel

As momentum and Energy is conserved

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

{\tfrac {1}{2}}m_{1}u_{1}^{2}+{\tfrac {1}{2}}m_{2}u_{2}^{2}={\tfrac {1}{2}}m_{1}v_{1}^{2}+{\tfrac {1}{2}}m_{2}v_{2}^{2}

From the two equations we get

v_{1}=\frac{m_1-m_2}{m_1+m_2}u_{1}+\frac{2m_2}{m_1+m_2}u_2\\\Rightarrow v_1=\frac{0.0025-0.005}{0.0025+0.005}\times 2.35+\frac{2\times 0.5}{0.4005+0.5}\times 0\\\Rightarrow v_1=-0.78333\ m/s

The final velocity of the penny is 0.78333 m/s in the opposite direction

v_{2}=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 0.0025}{0.0025+0.005}\times 2.35+\frac{0.005-0.0025}{0.005+0.0025}\times 0\\\Rightarrow v_2=1.566\ m/s

The final velocity of the nickel is 1.566 m/s in the same direction

6 0
3 years ago
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