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2 years ago

A 1000 Kg car traveling at 10 m/s hits the back of a 5000 Kg truck

Physics

1 answer:

insens350 [35]2 years ago

8 0

Answer:

0 m/s

The car becomes stationary

Explanation:

The law of conservation of linear momentum states that the sum of inital and final momentum should be equal and expressed as

$m_cu_c+m_tu_t=m_cv_c+m_tv_t$

Where m represent the mass, u and v are tge initial and final velocities while subscripts c and t represent car and truck.

Taking forward direction as positive then considering that the truck is originally at rest, we substitute original truck velocity with 0, mass of car and truck with 1000 kg and 5000 kg respectively then final truck velocity as 2 m/s as we take initial car velocity to be 10 m/s

1000*10+(5000*0)=5000*2+1000v

1000v=0

V=0

Therefore, the car finally becomes stationary.

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1 year ago

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6 months ago

What occurs when water droplets in the atmosphere become heavier than the air surrounding them can hold?

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2 years ago

Read 2 more answers

A particle with charge 3.20×10−19 c is placed on the x axis in a region where the electric potential due to other charges increa

lys-0071 [83]

Answer:

-5 V

Explanation:

The charged particle (which is positively charged) moves from point A to B, and its kinetic energy increases: it means that the particle is following the direction of the field, so its potential energy is decreasing (because it's been converted into potential energy), therefore it is moving from a point at higher potential (A) to a point at lower potential (B). This means that the value

vb−va

is negative.

We can calculate the potential difference between the two points by using the law of conservation of energy:

$\Delta K+ \Delta U=0\\\Delta K + q\Delta V=0$

where:

$\Delta K=+1.6\cdot 10^{-18} J$ is the change in kinetic energy of the particle

$q=3.2\cdot 10^{-19} C$ is the charge of the particle

$\Delta V =V_b-V_a$ is the potential difference

Re-arranging the equation, we can find the value of the potential difference:

$\Delta V=V_b-V_a = -\frac{\Delta K}{q}=-\frac{1.6\cdot 10^{-18} J}{3.2\cdot 10^{-19} C}=-5 V$

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