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suter [353]
3 years ago
12

The position of a particle as it moves along an y axis is given by y = (2.0cm)sin(πt/4), with t in second and y in centimeters.

Physics
1 answer:
irina [24]3 years ago
7 0

Part a)

At t = 0  the position of the object is given as

x = 0

At t = 2

x = 2 sin(\pi/2) = 2cm

so displacement of the object is given as

d = 2 - 0 = 2cm

so average speed is given as

v_{avg} = \frac{2}{2} = 1 cm/s

Part b)

instantaneous speed is given by

v = \frac{dy}{dt}

v = 2cos(\pi t/4 ) * \frac{\pi}{4}

now at t= 0

v = \frac{\pi}{2} cm/s

at t = 1

v = 2 cos(\pi/4) * \frac{\pi}{4}

v = \frac{\pi}{2\sqrt2}

at t = 2

v = 0

Part c)

Average acceleration is given as

a_{avg} = \frac{v_f - v_i}{t}

a_{avg} = \frac{0 - \frac{\pi}{2}}{2}

a = -\frac{\pi}{4} cm/s^2

Part d)

Now for instantaneous acceleration

As we know that

a =- \omega^2 y

at t = 0

a = -\frac{\pi^2}{16} * 0 = 0 cm/s^2

at t = 1

y = \sqrt2 cm

now we have

a = -\frac{\pi^2}{16}*\sqrt2

At t = 2 we have

y = 2 cm

a = -\frac{\pi^2}{16}*2

a = -\frac{\pi^2}{8}

<em>so above is the instantaneous accelerations</em>

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Gayle runs at a speed of 3.85 m/s and dives on a sled, initially at rest on the top of a frictionless snow-covered hill. After s
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Answer:

Final velocity at the bottom of hill is 15.56 m/s.

Explanation:

The given problem can be divided into four parts:

1. Use conservation of momentum to determine the speed of the combined mass (Gayle and sled)

From the law of conservation of momentum (perfectly inelastic collision), the combined velocity is given as:  

p_i = p_f  

m_1u_1 + m_2v_2 = (m_1 + m_2)v

v = \frac{(m_1u_1 + m_2v_2)}{(m_1 + m_2)}

v=\frac{[50.0\ kg)(3.85\ m/s) + 0]}{(50.0\ kg + 5.00\ kg)}= 3.5\ m/s  

2. Use conservation of energy to determine the speed after traveling a vertical height of 5 m.

The velocity of Gayle and sled at the instant her brother jumps on is found from the law of conservation of energy:  

E(i) = E(f)  

KE(i) + PE(i) = KE(f) + PE(f)  

0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)  

v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]}

Here, initial velocity is the final velocity from the first stage. Therefore:  

v(f) = \sqrt{[(3.5)^2+2(9.8)(5.00-0)]}= 10.5\ m/s

3. Use conservation of momentum to find the combined speed of Gayle and her brother.  

Given:

Initial velocity of Gayle and sled is, u_1(i)=10.5 m/s

Initial velocity of her brother is, u_2(i)=0 m/s

Mass of Gayle and sled is, m_1=55.0 kg

Mass of her brother is, m_2=30.0 kg

Final combined velocity is given as:

v(f) = \frac{[m_1u_1(i) + m_2u_2(i)]}{(m_1 + m_2)}  

v(f)=\frac{[(55.0)(10.5) + 0]}{(55.0+30.0)}= 6.79 m/s  

4. Finally, use conservation of energy to determine the final speed at the bottom of the hill.

Using conservation of energy, the final velocity at the bottom of the hill is:  

E(i) = E(f)  

KE(i) + PE(i) = KE(f) + PE(f)  

0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)  

v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]} \\v(f)=\sqrt{[(6.79)^2 + 2(9.8)(15 - 5.00)]}\\v(f)= 15.56\ m/s

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A cart of mass m = 0.12 kg moves with a speed v = 0.45 m/s on a frictionless air track and collides with an identical cart that
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Answer:

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The final kinetic energy of the system is expressed as;

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v is the final velocity of the bodies after collision

get the final velocity using the law of conservation of momentum

m1u1 + m2u2 = (m1+m2)v

0.12(0.45) + 0/12(0) = (0.12+0.12)v

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v = 0.054/0.24

v = 0.225m/s

Get the final kinetic energy;

KE = 1/2(m1+m2)v

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KE = 1/2(0.24)(0.050625)

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