Question

एक सुरक्षा परीक्षण में, द्रव्यमान 1000 किलो की एक कार ईंट की दीवार में चला जाता है।इसका बम्पर स्प्रिंग (क =5 x 10 एनएम) की तरह व्यवहार करता है और कार के विश्राम के समय 3 सेंटी मीटर की दूरी से संकुचित हो जाता है।कार की शुरुआती गति का निर्धारण करें। ​

Answer 1

Answer:

The initial speed of the car is 1.27 m/min

Explanation:

Mass of the car M = 1000 kg

spring constant of bumper that behaves like a spring k = 500 N/m

It is compressed by 3 cm

Therefore, x = 3 cm = 0.03 m

If the velocity of the car is v then from the question it is clear that the car comes to rest after the compression in spring is 3 cm

At that point, the whole kinetic energy of the car will be converted into potential energy of the spring

Therefore,

\frac{1}{2}Mv^2=\frac{1}{2}kx^2

2

1

Mv

2

=

2

1

kx

2

\implies Mv^2=kx^2⟹Mv

2

=kx

2

\implies 1000\times v^2=500\times (0.03)^2⟹1000×v

2

=500×(0.03)

2

\implies v^2=\frac{9\times 10^{-4}}{2}⟹v

2

=

2

9×10

−4

\implies v=\frac{3\times 10^{-2}}{\sqrt{2}}⟹v=

2

3×10

−2

\implies v=2.12\times 10^{-2}⟹v=2.12×10

−2

m/s

\implies v=2.12\times 10^{-2}\times 60⟹v=2.12×10

−2

×60 m/min

\implies v=1.27⟹v=1.27 m/min

Therefore, the initial speed of the car is 1.27 m/min

Hope this answer is helpful.