Question

If a car moves with a velocity of 10 m/s after applying brakes comes to rest in 5 seconds.

Calculate (i) acceleration and (ii) change in momentum if the car weighs 800 kg.
Please can you help me fast it is my test !!​

Answer 1

Answer:

1. 2 m/s^2

2. 8000 kgm/s

Explanation:

Acceleration = Velocity / Time

Momentum = Mass × Velocity

Answer 2

Provided that:

• Initial velocity = 10 m/s
• Final velocity = 0 m/s
• Time taken = 5 seconds
• Mass of car = 800 kg

Don't be confused! Final velocity cames as zero because it is given that breaks are applied to moving car to brings it at rest.

To calculate:

• Acceleration
• Change in momentum

Solution:

• Acceleration = -2 m/s sq.

• Change in momentum = -8000 kg m/s

Using concepts:

• Acceleration formula
• Change in momentum formula

Using formulas:

Acceleration is given by

• ${\small{\underline{\boxed{\sf{a \: = \dfrac{v-u}{t}}}}}}$

Change in momentum is given by

• ${\small{\underline{\boxed{\sf{\Delta p \: = m(v-u)}}}}}$

Or

• ${\small{\underline{\boxed{\sf{\Delta p \: = mv \: - mu}}}}}$

Where, a denotes acceleration, v denotes final velocity, u denotes initial velocity, t denotes time taken, m denotes mass, mg denotes final momentum, mu denotes initial momentum, p denotes momentum and ∆p denotes change in momentum

Required solution:

~ Firstly let us calculate acceleration!

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{0-10}{5} \\ \\ :\implies \sf a \: = \dfrac{-10}{5} \\ \\ :\implies \sf a \: = \cancel{\dfrac{-10}{5}} \: (Cancelling) \\ \\ :\implies \sf a \: = -2 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = -2 \: ms^{-2} \\ \\ :\implies \sf Retardation \: = -2 \: ms^{-2}$

Therefore, acceleration = -2 metre per second sq. It came in negative because breaks are applied! That's why acceleration became retardation!

~ Now let us calculate change in momentum by using suitable formula(s)

By method first, let us calculate change in momentum...

$:\implies \sf \Delta p \: = m(v-u) \\ \\ :\implies \sf \Delta p \: = 800(0-10) \\ \\ :\implies \sf \Delta p \: = 800(-10) \\ \\ :\implies \sf \Delta p \: = -8000 \: kg \: ms^{-1} \\ \\ :\implies \sf Change \: in \: momentum \: = -8000 \: kg \: ms^{-1}$

By second method, let us calculate change in momentum...

$:\implies \sf \Delta p \: = mv \: - mu \\ \\ :\implies \sf \Delta p \: = 800 \times 0 - 800 \times 10 \\ \\ :\implies \sf \Delta p \: = 0 - 8000 \\ \\ :\implies \sf \Delta p \: = -8000 \: kg \: ms^{-1}$