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3 years ago

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A 2 kg ball of putty moving to the right at 3m/s has a head-on inelastic collision with a 2 kg ball of putty at rest. What is th

e final magnitude and direction of the velocity of the stuck together balls after the collision?

1 answer:

sleet_krkn [62]3 years ago

3 0

Answer:1.2 m/s

Explanation:

mass of ball $`(m_1)=2 kg$ `

initial Velocity of $m_1=3 m/s$ `

mass of another ball $(m_2)=2 kg$ `

initial velocity of $m_2=0$ `

Conserving momentum

$m_1u_1+m_2u_2=(m_1+m_2)v$ `

$2\times 3+2\times 0=(2+3)v$ `

$v=\frac{6}{5}=1.2 m/s$ `

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A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7

aleksandrvk [35]

(a) 1440.5 Hz

The general formula for the Doppler effect is

$f'=(\frac{v+v_r}{v+v_s})f$

where

f is the original frequency

f is the apparent frequency

$v$ is the velocity of the wave

$v_r$ is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

$v_s$ is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

f = 1110 Hz

v = 334 m/s

In the reflector frame (= on surface B), we have also

$v_s = v_A = -28.7 m/s$ (surface A is the source, which is moving towards the receiver)

$v_r = +62.2 m/s$ (surface B is the receiver, which is moving towards the source)

So, the frequency observed in the reflector frame is

$f'=(\frac{334 m/s+62.2 m/s}{334 m/s-28.7 m/s})1110 Hz=1440.5 Hz$

(b) 0.232 m

The wavelength of a wave is given by

$\lambda=\frac{v}{f}$

where

v is the speed of the wave

f is the frequency

In the reflector frame,

f = 1440.5 Hz

So the wavelength is

$\lambda=\frac{334 m/s}{1440.5 Hz}=0.232 m$

(c) 1481.2 Hz

Again, we can use the same formula

$f'=(\frac{v+v_r}{v+v_s})f$

In the source frame (= on surface A), we have

$v_s = v_B = -62.2 m/s$ (surface B is now the source, since it reflects the wave, and it is moving towards the receiver)

$v_r = +28.7 m/s$ (surface A is now the receiver, which is moving towards the source)

So, the frequency observed in the source frame is

$f'=(\frac{334 m/s+28.7 m/s}{334 m/s-62.2 m/s})1110 Hz=1481.2 Hz$

(d) 0.225 m

The wavelength of the wave is given by

$\lambda=\frac{v}{f}$

where in this case we have

v = 334 m/s

f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

$\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m$

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