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Anestetic [448]
4 years ago
12

(96 Points )A bus and a bicycle have a head-on collision. Compare the force of impact between the bus and the bicycle; compare t

he accelerations of the bus and bicycle.
Physics
1 answer:
lidiya [134]4 years ago
6 0

Answer:

Small acceleration for the bus and major for the bicycle

Explanation:

In a head-on collision of this type, i order to find force involve in a collision, one has to study the change in linear momentum of each object. Because Force s defined as the change of momentum with time.

Recall that momentum is the product of the mass of the object times its velocity. The mass of the bus (M) is much larger than that of the bicycle (m), and even if their speeds are similar, the momentum of the bus will be much larger than that of the bike due to this mass difference. We could assume an inelastic collision (where the two objects stay together after the collision),

The initial total momentum of the system (imagining that the bus is pointing to the right (positive convention), while the bike is pointing in the opposite (to the left), can be written as: Pi=Mv-mv (we are assuming that both vehicles go at the same speed for simplicity)

The final total momentum of the system that underwent an inelastic collision will be: P_f=(M+m)v_f

Therefore, since the mass M is much larger than mass m, the final momentum will be pretty similar to the initial momentum of the bus.

The change in momentum for the bus in the fraction of time of the collision will be almost unnoticeable for the bus (minimal change in its momentum, and therefore minimal acceleration accounting for the change).

For the bicycle, the change in momentum involves change in the direction of motion, (going to the left before the collision, and to the right afterwards) which will include a mayor acceleration backwards.

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Answer:

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If the baseball and the plastic ball were moving at the same speed which ball would hit a bat harder
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A motorcycle is following a car that is traveling at a constant speed on a straight highway. Initially, the car and the motorcyc
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Answer:

(a) 3.807 s

(b) 145.581 m

Explanation:

Let Δt = t2 - t1 be the time it takes from the moment when the motorcycle starts to accelerate until it catches up with the car. We know that before the acceleration, both vehicles are travelling at a constant speed. So they would maintain a distance of 58 m prior to the acceleration.

The distance traveled by car after Δt (seconds) at v_c = 23m/s speed is

s_c = \Delta t v_c = 23\Delta t

The distance traveled by the motorcycle after Δt (seconds) at m_m = 23 m/s speed and acceleration of a = 8 m/s2 is

s_m = \Delta t v_m + a\Delta t^2/2

s_m = 23\Delta t + 8\Delta t^2/2 = 23 \Delta t + 4 \Delta t^2

We know that the motorcycle catches up to the car after Δt, so it must have covered the distance that the car travels, plus their initial distance:

s_m = s_c + 58

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4 \Delta t^2 = 58

\Delta t^2 = 14.5

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(b)

s_m = 23 \Delta t + 4 \Delta t^2

s_m = 23*3.807 + 58 = 145.581 m

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