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  • Using second law of motion derive the relation between force and acceleration A bullet of 10 g strikes a sand bag at a speed of 10^3 ms ^{-1} and gets embedded after travelling 5 cm Calculate

Using second law of motion, derive the relation between force and acceleration. A bullet of   10 g strikes a sand-bag at a speed of 103 ms-1 and gets embedded after travelling 5 cm. Calculate

(i) the resistive force exerted by the sand on the bullet

(ii) the time taken by the bullet to come to rest.

Answers (1)

By Newton’s second law, we know that force on a body will be equal to the product of mass and acceleration.

F = ma

Now as the moving bullet with the speed 103 m/s, get embedded into the sandbag and stops. It means that the speed of bullet changes. Hence, it must have some acceleration.

By using the third equation of motion: we can find out the acceleration as we know the initial velocity, final velocity and displacement.

v^2 =u^2+2as\\wherev=0;u=10^3 m/sands=0.05m\\0^2 =10^6+2a(0.05)\\a=-10^7 m/s^2

Hence the force will be:

F=ma=10^{-2}\times10^7 N=10^5 N
By the first equation of motion:

v=u+at\\Whereu=10^3;a=-10^7;v=0\\\Rightarrow t=\frac{10^3}{10^7}=10^{-4} sec

 

Posted by

Gurleen Kaur

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