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Ahat [919]
3 years ago
11

In an elastic collision, a 580 kg bumper car collides directly from behind with a second, identical bumper car that is traveling

in the same direction. The initial speed of the leading bumper car is 1.50 m/s and that of the trailing car is 2.75 m/s. Assuming that the mass of each driver is 65 kg, what are the cars' final speeds after the collision?'
Physics
1 answer:
kozerog [31]3 years ago
8 0

Answer:  vl = 2.75 m/s vt = 1.5 m/s

Explanation:

If we assume that no external forces act during the collision, total momentum must be conserved.

If both cars are identical and also the drivers have the same mass, we can write the following:

m (vi1 + vi2) = m (vf1 + vf2) (1)

The sum of the initial speeds must be equal to the sum of the final ones.

If we are told that kinetic energy must be conserved also, simplifying, we can write:

vi1² + vi2² = vf1² + vf2² (2)

The only condition that satisfies  (1) and (2) simultaneously is the one in which both masses exchange speeds, so we can write:

vf1 = vi2 and vf2 = vi1

If we call v1 to the speed of the leading car, and v2 to the trailing one, we can finally put the following:

vf1 = 2.75 m/s  vf2 = 1.5 m/s

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A 2.2 kgkg block slides along a frictionless surface at 1.2 m/sm/s . A second block, sliding at a faster 4.0 m/sm/s , collides w
aleksklad [387]

Answer:

0.6kg

Explanation:

the unknown here is the mass of the second block

applying the law of the conservation of momentum

m₁v₁ + m₂v₂ = (m₁ + m₂) v₃

where m₁=mass of first block=2.2kg

m₂=mass of colliding block= ?

v₁= velocity of first block=1.2m/s

v₂=velocity of colliding block=4.0m/s

v₃= final velocity of combined block=1.8m/s

applying the formula above

(2.2 × 1.2) + (m₂ × 4) = (2.2 + m₂) × 1.8

2.64 + 4m₂ = 3.96 + 1.8m₂

collecting like terms

4m₂ - 1.8m₂ = 3.96 - 2.64

2.2m₂=1.32

divide both sides by 2.2

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4 0
3 years ago
In one of the original Doppler experiments, a tuba was played at a frequency of 64.0 Hz on a moving flat train car, and a second
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Answer:

 f_{beat} = 1.64\ Hz

Explanation:

given,

frequency of tuba.f = 64 Hz

Speed of train approaching, v = 8.50 m/s

beat frequency = ?

using Doppler's effect formula

 f' = f(\dfrac{v}{v-v_s})

v_s is the velocity of the source

v is the speed of sound, v = 340 m/s

now,

 f' = 64\times (\dfrac{340}{340 - 8.50})

       f' = 65.64 Hz

now, beat frequency is equal to

 f_{beat} = f' - f

 f_{beat} = 65.64 - 64

 f_{beat} = 1.64\ Hz

hence, beat frequency is equal to 1.64 Hz

3 0
3 years ago
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