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Anna35 [415]
3 years ago
7

After a hold up, Robin Banks flees in his 1575 kg getaway car at 20 m/s. He crashes into a 45 kg highway barrel which is at rest

. If Robin Bank’s car moves at 18.9 m/s after the collision, how fast does the barrel move after being hit?
Physics
2 answers:
fiasKO [112]3 years ago
7 0

Answer:

v = 38.9 m/s

Explanation:

Dimas [21]3 years ago
6 0

v = 38.5 m/s

Explanation:

P1 = P2 (conservation law of linear momentum)

(1575 kg)(20 m/s) + (45 m/s)(0) = (1575 kg)(18.9 m/s) + 45v

Solving for v

v = 38.5 m/s

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If a body having mass 40kg started moving initially with rest and it takes a velocity of 20m/sec in time 4 seconds. Find the val
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{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Mass \ of \ the \ body \ (m) = 40 \ kg}

\:\:\:\:\bullet\:\:\:\sf{Final \ velocity \ of \ the \  body \ (v) = 20 \ m/s}

\:\:\:\:\bullet\:\:\:\sf{Initial \ velocity \ of \ the \  body \ (u) = 0}

\\

{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Force \ exerted \ by \ the \  body \ ( F)}

\\

{\mathfrak{\underline{\purple{\:\:\: Solution:-\:\:\:}}}} \\ \\

☯ <u>Using 1st equation of motion </u>

\\

\dashrightarrow\:\: \sf{v = u + at}

\\

\dashrightarrow\:\: \sf{20 = 0 + a(4)}

\\

\dashrightarrow\:\: \sf{20 = 4a}

\\

\dashrightarrow\:\: \sf{\dfrac{\cancel{20}}{\cancel{4}} = a}

\\

\dashrightarrow\:\: \sf{a = 5}

\\

☯ <u>Now, Finding the force exerted </u>

\\

\dashrightarrow\:\: \sf{F = ma}

\\

\dashrightarrow\:\: \sf{F = 40 \times 5}

\\

\dashrightarrow\:\: \sf{F = 200 \ N}

\\

☯ <u>Hence</u>, \\

\:\:\:\:\star\:\:\:\sf{The \ force \ exerted \ by \ the \ body \ is \ 200N}

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