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katrin2010 [14]

3 years ago

11

A truck with a mass of 1650 kg and moving with a speed of 15.0 m/s rear-ends a 779 kg car stopped at an intersection. The collis

ion is approximately elastic since the car is in neutral, the brakes are off, the metal bumpers line up well and do not get damaged. Find the speed of both vehicles after the collision in meters per second.

1 answer:

ivanzaharov [21]3 years ago

6 0

Answer:

Truck's speed = 5.21 m/s

Car's speed = 20.2 m/s

Explanation:

Given:

Mass of truck = M = 1650 kg

Speed of the truck initially = U = 15 m/s

Mass of the car = m = 779 kg

Initial speed of the car =u = 0

From the momentum conservation, Total initial momentum = Total final momentum.

M V+m U = M V +m v

⇒ (1650)(15) + 779×0 = (1650)V + 779 v

⇒ 24750 = 1650 V+779 v →(1)

Since the collision is elastic, relative velocity of approach = relative velocity of separation. 15 = v - V

⇒ v =V + 15; This is now substituted in the equation(1) above.

24750 = 1650 V + (799) (V+15)

⇒ 24750 = 1650 V + 799 V + 11985

⇒ 2449 V = 12765

⇒ Final velocity of the truck = $\frac{12765}{2449}$ = 5.21 m/s

Final velocity of the car = v = V+15 = 5.21 + 15 = 20.2 m/s

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Two loudspeakers emit sound waves of the same frequency along the x-axis. The amplitude of each wave is a. The sound intensity i

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Answer:

Explanation:

To find the amplitude of the sound, we must first determine the wavelength and the phase difference between the two speakers.

For the wavelength;

Recall that, the separation between two successive max. and min. intensity points are $\dfrac{\lambda}{2}$

Thus; for both speakers; the wavelength of the sound is:

$\dfrac{\lambda}{2} = (10+30) cm$

$\dfrac{\lambda}{2} = (40) cm$

λ = 80 cm

The relation between the path difference(Δx) and the phase difference(Δ∅) is:

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

where;

Δx = 10 cm

λ = 80 cm

Δ∅ = π rad

∴

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

$\pi \ rad = \dfrac{2 \pi}{80 \ cm}(10 \ cm) + \Delta \phi_o$

$\pi \ rad = \dfrac{2 \pi}{8}+ \Delta \phi_o$

$\pi \ rad = \dfrac{ \pi}{4}+ \Delta \phi_o$

$\Delta \phi_o = \pi -\dfrac{ \pi}{4}$

$\Delta \phi_o = \dfrac{ 4\pi - \pi}{4}$

$\Delta \phi_o = \dfrac{ 3\pi}{4} \ rad$

Suppose both speakers are placed side-by-side, then the path difference between the two speakers is: Δx = 0 cm

Thus, we have:

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

$\Delta \phi = \dfrac{2 \pi}{\lambda}(0 \ cm ) + \dfrac{3 \pi}{4} \ rad$

$\Delta \phi = \dfrac{3 \pi}{4} \ rad$

∴

The amplitude of the sound wave if the two speakers are placed side-by-side is:

$A = 2a \ cos \bigg (\dfrac{\Delta \phi }{2} \bigg)$

$A = 2a \ cos \bigg (\dfrac{\dfrac{3 \pi}{4} }{2} \bigg)$

$A = 2a \ cos \bigg ({\dfrac{3 \pi}{8} } \bigg)$

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