CBSE
Gujarat Board
Haryana Board
Class 10
Class 12
The figure shows the position –time (x – t) graph of one–dimensional motion of a body of mass 0.4 kg. The magnitude of each impulse is
0.4 (Ns)
0.8 Ns
1.6 Ns
0.2 Ns
B.
0.8 Ns
From the graph it is a straight line, so uniform motion. Because of impulse direction of velocity changes as can be seen from the slope of the graph.
(∴ impulse (l) = f x t
⇒ I = mat
Initial velocity, v_{1} = 2/2 = 1 ms^{-1}
Final velocity v_{2} = 2/2 = - 1 ms^{-1}
p_{i} = mv1 = 0.4 N-s
p_{f} = mv2 = -0.4 N-s
J = p_{f}-p_{i} = -0.4-0.4
= - 0.8 N-s
|J| = 0.8 N-s
From the graph it is a straight line, so uniform motion. Because of impulse direction of velocity changes as can be seen from the slope of the graph.
(∴ impulse (l) = f x t
⇒ I = mat
Initial velocity, v_{1} = 2/2 = 1 ms^{-1}
Final velocity v_{2} = 2/2 = - 1 ms^{-1}
p_{i} = mv1 = 0.4 N-s
p_{f} = mv2 = -0.4 N-s
J = p_{f}-p_{i} = -0.4-0.4
= - 0.8 N-s
|J| = 0.8 N-s