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Bad White [126]
3 years ago
5

You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction between

your feet and the ice. A friend throws you a ball of mass 0.400 kg that is traveling horizontally at 11.5 m/s. Your mass is 75.0 kg.
(A) If you catch the ball, with what speed do you and the ball move afterwards?
(B) If the ball hits you and bounces off your chest, so that afterwards it is moving horizontally at 7.80m/s in the opposite direction, what is your speed after the collision?
Physics
1 answer:
9966 [12]3 years ago
3 0

Answer:

(a) 0.061 m/s

(b) 0.103 m/s

Explanation:

From the law of conservation of momentum, the sum of initial momentum equals the sum of final momentum

Momentum, p=mv where m is the mass and v is the velocity

m_1v_1=(m_1+m_2)v_c where v_c is the common velocity, v_1 is the velocity of the ball m_1 and m_2 are masses of the ball and person respectively

Substituting the given values then

0.400\times 11.5 = (0.400+75)v_c\\v_c=0.061007958\approx 0.061 m/s

(b)

Momentum, p=mv where m is the mass and v is the velocity

m_1v_1=m_1v_2+m_2v_3

v_1 is the velocity of the ball , v_2 is the velocity of ball afterwards and v_3 is your speed, m_1 and m_2 are masses of the ball and person respectively. Since it bounces back, we give it a negative value hence

0.400\times 11.5= 0.4\times -7.8+75v_3\\v_3=0.102933333\approx 0.103 m/s

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. A 13-g goldfinch has a speed of 8.5 m/s. What is its kinetic energy?
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In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the following expression,
eduard

(a) 4.06 cm

In a simple harmonic motion, the displacement is written as

x(t) = A cos (\omega t + \phi) (1)

where

A is the amplitude

\omega is the angular frequency

\phi is the phase

t is the time

The displacement of the piston in the problem is given by

x(t) = (5.00 cm) cos (5t+\frac{\pi}{5}) (2)

By putting t=0 in the formula, we find the position of the piston at t=0:

x(0) = (5.00 cm) cos (0+\frac{\pi}{5})=4.06 cm

(b) -14.69 cm/s

In a simple harmonic motion, the velocity is equal to the derivative of the displacement. Therefore:

v(t) = x'(t) = -\omega A sin (\omega t + \phi) (3)

Differentiating eq.(2), we find

v(t) = x'(t) = -(5 rad/s)(5.00 cm) sin (5t+\frac{\pi}{5})=-(25.0 cm/s) sin (5t+\frac{\pi}{5})

And substituting t=0, we find the velocity at time t=0:

v(0)=-(25.00 cm/s) sin (0+\frac{\pi}{5})=-14.69 cm/s

(c) -101.13 cm/s^2

In a simple harmonic motion, the acceleration is equal to the derivative of the velocity. Therefore:

a(t) = v'(t) = -\omega^2 A cos (\omega t + \phi)

Differentiating eq.(3), we find

a(t) = v'(t) = -(5 rad/s)(25.00 cm/s) cos (5t+\frac{\pi}{5})=-(125.0 cm/s^2) cos (5t+\frac{\pi}{5})

And substituting t=0, we find the acceleration at time t=0:

a(0)=-(125.00 cm/s) cos (0+\frac{\pi}{5})=-101.13 cm/s^2

(d) 5.00 cm, 1.26 s

By comparing eq.(1) and (2), we notice immediately that the amplitude is

A = 5.00 cm

For the period, we have to start from the relationship between angular frequency and period T:

\omega=\frac{2\pi}{T}

Using \omega = 5.0 rad/s and solving for T, we find

T=\frac{2\pi}{5 rad/s}=1.26 s

4 0
2 years ago
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