A lead bullet of mass 0.05kg is fired with a velocity of ...
A lead bullet of mass 0.05kg is fired with a velocity of 200ms\(^{-1}\) into a lead block of mass 0.95kg. Given that the lead block can move freely, the final kinetic energy after impact is
50 J
100 J
150 J
200 J
Correct answer is A
From principle of conservation of linear momentum,
(0.05 x 200) + (0.95 x 0) = (0.05 + 0.95) x V (since collision is inelastic).
10 + 0 = V
Thus V = 10m/s.
Recall Kinetic Energy = \(\frac{1}{2} mv^2\)
\(\therefore\) K.E = 1/2 (0.05 + 0.95) x 10\(^2\)
K.E = 1/2 (1 x 100) = 50 J.
Radio wave Micro wave Infrared wave P Q R Gamma Ray The table above shows the com...
The force per unit charge experienced at a point in a field is the ...
When the brakes in a car are applied, the frictional force force on the tyres is ...
Which of the following statements is NOT true of \(\gamma\)-rays?...
Which of the following is not a part of an A.C generator? ...
Which of the following statements about the concept of solid friction is NOT true? ...
In the formation of sea breeze. wind blows from ...
The heat capacity of a substance is the energy ...
The image formed by a concave mirror is real, inverted and magnified, when object is placed ...