Question

A motorcar of mass 1200 kg is moving along
a straight line
with a uniform velocity of 90 km/h. Its
velocity is slowed down
to 18 km/h in 4 s by an unbalanced external

force. Calculate

the acceleration and change in momentum.

Also calculate the

magnitude of the force required.

spammers immediately reported....


​

Answer 1

Given :-

• Mass of motorcar, m = 1200 kg

• Initial velocity of motorcar, u = 90 kmh⁻¹ = $\sf{90\times\dfrac{5}{18}\;\;kmh^{-1}}$ = 25 ms⁻¹

• Final velocity of motorcar, v = 18 kmh⁻¹ =$\sf{18\times\dfrac{5}{18}\;\;kumhar^{-1}}$= 5 ms⁻¹

• Time in which motorcar is slowed down, t = 4 s

To find :-

• Acceleration of the Motorcar, a =?

• Change in momentum, Δ p =?

• Magnitude of force for slowing down the motorboat, F =?

Formulae required :-

the First equation of motion

• $\red{\bigstar}\boxed{\sf{v=u+at}}$

Formula to calculate change in momentum

• $\red{\bigstar}\boxed{\sf{\Delta\;p=p_f-p_i=mv-mu}}$

Formula to calculate Force required

• $\red{\bigstar}\boxed{\sf{F=\dfrac{\Delta\;p}{t}}}$

• [ Where v is final velocity, u is initial velocity, a is acceleration, t is time taken, Δ p is change in momentum, $\sf{p_i}$ is initial momentum, $\sf{p_f}$ is final momentum, F is force required and m is mass of body ]

Solution :-

Calculating acceleration of motorboat

Using first equation of motion

• $\implies\sf{5=25+a\times 4}$
• $\implies\sf{4a=-20}$
• $\implies\underline{\underline{\red{\sf{a=-5\;\;ms^{-2}}}}}$

Calculating change in momentum

Using formula for change in momentum

• $\implies\sf{\Delta\;p=p_f-p_i=mv-mu}$
• $\implies\sf{\Delta\;p=1200\times 5 - 1200 \times 25}$
• $\implies\sf{\Delta\;p=1200\times 5 - 1200 \times 25}$
• $\implies\underline{\underline{\red{\sf{\Delta\;p=-24000\;\;Kgms^{-1}}}}}$

Calculating the magnitude of force required

Using formula for calculating force

• $\implies\sf{F=\dfrac{\Delta\;p}{t}}$
• $\implies\sf{F=\dfrac{-24000}{4}}$
• $\implies\underline{\underline{\red{\sf{F=-6000\;\;J}}}}$

Therefore,

• Acceleration of motorboat is -5 m/s².
• Momentum is reduced by 24000 kg ms⁻¹.
• Magnitude of force required is 6000 J.

Answer 2

Step-by-step explanation:

Given :-

Mass of motorcar, m = 1200 kg

Initial velocity of motorcar, u = 90 kmh⁻¹ = \sf{90\times\dfrac{5}{18}\;\;kmh^{-1}}90×

18

5

kmh

−1

= 25 ms⁻¹

Final velocity of motorcar, v = 18 kmh⁻¹ =\sf{18\times\dfrac{5}{18}\;\;kumhar^{-1}}18×

18

5

kumhar

−1

= 5 ms⁻¹

Time in which motorcar is slowed down, t = 4 s

To find :-

Acceleration of the Motorcar, a =?

Change in momentum, Δ p =?

Magnitude of force for slowing down the motorboat, F =?

Formulae required :-

the First equation of motion

\red{\bigstar}\boxed{\sf{v=u+at}}★

v=u+at

Formula to calculate change in momentum

\red{\bigstar}\boxed{\sf{\Delta\;p=p_f-p_i=mv-mu}}★

Δp=p

f

−p

i

=mv−mu

Formula to calculate Force required

\red{\bigstar}\boxed{\sf{F=\dfrac{\Delta\;p}{t}}}★

F=

t

Δp

[ Where v is final velocity, u is initial velocity, a is acceleration, t is time taken, Δ p is change in momentum, \sf{p_i}p

i

is initial momentum, \sf{p_f}p

f

is final momentum, F is force required and m is mass of body ]

Solution :-

Calculating acceleration of motorboat

Using first equation of motion

\implies\sf{5=25+a\times 4}⟹5=25+a×4

\implies\sf{4a=-20}⟹4a=−20

\implies\underline{\underline{\red{\sf{a=-5\;\;ms^{-2}}}}}⟹

a=−5ms

−2

Calculating change in momentum

Using formula for change in momentum

\implies\sf{\Delta\;p=p_f-p_i=mv-mu}⟹Δp=p

f

−p

i

=mv−mu

\implies\sf{\Delta\;p=1200\times 5 - 1200 \times 25}⟹Δp=1200×5−1200×25

\implies\sf{\Delta\;p=1200\times 5 - 1200 \times 25}⟹Δp=1200×5−1200×25

\implies\underline{\underline{\red{\sf{\Delta\;p=-24000\;\;Kgms^{-1}}}}}⟹

Δp=−24000Kgms

−1

Calculating the magnitude of force required

Using formula for calculating force

\implies\sf{F=\dfrac{\Delta\;p}{t}}⟹F=

t

Δp

\implies\sf{F=\dfrac{-24000}{4}}⟹F=

4

−24000

\implies\underline{\underline{\red{\sf{F=-6000\;\;J}}}}⟹

F=−6000J

Therefore,

Acceleration of motorboat is -5 m/s².

Momentum is reduced by 24000 kg ms⁻¹.

Magnitude of force required is 6000 J.