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Rama09 [41]
3 years ago
8

A 6.00 kg object at rest is suddenly broken apart into two fragments by an explosion. The fragment with mass 3.50 kg is moving 2

4.0m/s to the left. The other fragment is moving _______m/s to the right. Group of answer choices
Physics
1 answer:
nexus9112 [7]3 years ago
6 0

Answer: 33.6 m/s

Explanation:

Given

The initial mass of the object is 6\ kg

It breaks into two fragments of 3.5 and 2.5 kg

3.5 kg part moves at 24 m/s towards left

Suppose 2.5 kg part moves with a speed of v towards the right

As the center of mass is at rest before the collision, it continues to remain at rest after the collision.

\therefore 6\times 0=3.5\times 24-2.5\times v\\\Rightarrow 3.5\times 24=2.5\times v\\\\\Rightarrow v=33.6\ m/s

Therefore, the velocity of 2.5 kg mass is 33.6 m/s towards right.

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The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance D_{A} beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed v_{A}. Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed v_{B}, which is greater than v_{A}.

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Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

x=x_{0}+vt

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The equation will be:

x_{A}=D_{A}+v_{A}t

Car B started at the starting line. So, its equation is

x_{B}=v_{B}t

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D_{A}+v_{A}t=v_{B}t

v_{B}t-v_{A}t=D_{A}

t(v_{B}-v_{A})=D_{A}

t=\frac{D_{A}}{v_{B}-v_{A}}

Car B meet with Car A after t=\frac{D_{A}}{v_{B}-v_{A}} units of time.

Part B: With the meeting time, we can determine the position they will be:

x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )

x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}

Since Car B started at the starting line, the distance Car B will be when it passes Car A is x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}} units of distance.

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