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Ann [662]
2 years ago
9

in an automobile crash, a vehicle that was stopped at a red light is rear-ended by another vehicle. The vehicles have the same m

ass. If the tire marks show that the two vehicles moved after the collision at 4 m/s, what was the speed of the vehicle before the collision
Physics
1 answer:
baherus [9]2 years ago
6 0

The initial speed of the vehicle before the collision is 8 m/s.

  • Let the mass of the vehicle = m
  • Let the initial speed of the vehicle stopped = u
  • The initial speed of the vehicle parked at the red light = 0

<h3>Principle of conservation of linear momentum</h3>
  • The initial speed of the vehicle before the collision is calculated by applying principle of conservation of linear momentum as follows;

m_1 u_1 + m_2 u_2 = v(m_1 + m_2)\\\\mu + m(0) = 4(m+ m)\\\\mu = 4(2m)\\\\mu = 8m\\\\u = 8 \ m/s

Thus, the initial speed of the vehicle before the collision is 8 m/s.

Learn more about inelastic collision here: brainly.com/question/7694106

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M (b) Calculate the speed of an electron that is accelerated through the same electric potential difference.
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The speed of a electron that is accelerated from rest through an electric potential difference of 120 V  is 6.49\times10^6m/s

<h3>How to calculate the speed of the electron?</h3>

We know, that the  energy of the system is always conserved.

Using the Law of Conservation of energy,

\triangle K+\triangle U=0

Here, K is the kinetic energy and U is the potential energy.

Now, substituting the formula of U and K, we get:

(\frac{1}{2} mv^2)+(-q\triangle V) =0------(1)

Here,

m is the mass of the electron

v is the speed of the electron

q is the charge on the electron

V is the potential difference

Let v_f and v_i represent the final and initial speed.

Here, v_i =0

Solving for v_f, we get:

v_f=\sqrt{\frac{-2q\triangle V}{m} }

=\sqrt{\frac{2\times1.602\times10^{-19}\times 120}{9.109\times10^{-31}} }

=6.49\times10^6m/s

To learn more about the conservation of energy, refer to:

brainly.com/question/2137260

#SPJ4

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