UPSEE 2017 Solved Paper

Given, mass of block (m)=1kg
Velocity of block at time (t)2s is −3m∕s
Impulse during t=2s to t=4sFrom the graph,
I = Area of ΔABC + Area of ΔCDE
=

1

2

(3−2)×5+

1

2

(4−3)(−5)
=

1

2

×1×5+

1

2

×(1)(−5)
=

5

2

−

5

2

=0
We know that,
Impulse = Change in momentum
0=mvf−mvi
⇒ mvf−mvi=0
⇒ mvf=mvi
vf=vi=−3m∕s or |3m∕s|