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ikadub [295]
3 years ago
6

A red car has a head-on collision with an approaching blue car with the same magnitude of momentum. A green car driving with the

same momentum as the other cars collides with an enormously massive wall. Which of the three cars will experience the greatest impulse
Physics
2 answers:
WARRIOR [948]3 years ago
5 0

Answer:

The head on Collison because you have both cars going (for the sake of it 30mph) and they both collide the energy from that is 60mph because the speed is combined with the two cars.

Marat540 [252]3 years ago
4 0

All three cars experience the same impulse.

Impulse is equal to change in momentum.

Each car starts with the same amount of momentum and ends up with zero, so the magnitudes of all three changes are equal.

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Explanation:

A) To prove the motion of the center of mass of the cylinders is simple harmonic:

System diagram for given situation is shown in attached Fig. 1

We can prove the motion of the center of mass of the cylinders is simple harmonic if

a_{x} = -\omega^{2}  x

where aₓ is acceleration when attached cylinders move in horizontal direction:

<h3>PROOF:</h3>

rotational inertia for cylinders  is given as:

                                  I=\frac{1}{2}MR^{2} -----(1)

Newton's second law for angular motion is:

                                             ∑τ = Iα ------(2)

For linear motion in horizontal direction it is:

                                             ∑Fₓ = Maₓ ------ (3)

By definition of torque:

                                               τ  = RF --------(4)        

Put (4) and (1) in (2)

                                       RF=\frac{1}{2}MR^{2}\alpha

                                       RF=\frac{1}{2}MR^{2}\alpha

from Fig 3 it can be seen that fs is force by which the cylinders roll without slipping as they oscillate

So above equation becomes

                                   f_{s}=\frac{1}{2}MR\alpha------ (5)

As angular acceleration is related to linear by:

                                          a= R\alpha

Eq (5) becomes

                                    f_{s}=\frac{1}{2}Ma_{x}---- (6)

aₓ shows displacement in horizontal direction

From (3)

                                              ∑Fₓ = Maₓ

Fₓ is sum of fs and restoring force that spring exerts:

                                  \sum F_{x} = f_{s} - kx ----(7)

Put (7) in (3)

                                  f_{s} - kx  = Ma_{x}[/tex] -----(8)

Using (6) in (8)

                               \frac{1}{2}Ma_{x} - kx =Ma_{x}

                                     a_{x} = \frac{2k}{3M} x --- (9)

For spring mass system

                                  a= -\omega^{2} x ----- (10)

Equating (9) and (10)

                                  \omega^{2} = \frac{2k}{3M}

\omega = \sqrt{ \frac{2k}{3M}}

then (9) becomes

                                a_{x} = - \omega^{2}x

(The minus sign says that x and  aₓ  have opposite directions as shown in fig 3)

This proves that the motion of the center of mass of the cylinders is simple harmonic.

<h3 /><h3>B) Time Period</h3>

Time period is related to angular frequency as:

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

                                  T = 2\pi \sqrt{\frac{3M}{2k}

                           

 

5 0
3 years ago
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