VITEEE 2009 Paper

Momentum, p = m.v ⇒ v = (

p

m

)
Kinetic energy, KE =

1

2

mv2=

1

2

m(

p2

m2

)=

1

2m

p2
or, KE∝p2‌‌‌‌(∵

1

2m

=constant)
Hence, the graph between KE and p2 will be linear.
Now, kinetic energy KE =

1

2

mv2
The velocity component at point P,
vy=(u‌sin‌θ−gt) and vx=(u‌cos‌θ)
Resultant velocity at point P,

→

v

=vy

^

j

+vx

^

i

=(u‌sin‌θ−gt)

^

j

+u‌cos‌θ‌

^

i

|

→

v

|=√(u‌cos‌θ)2+(u‌sin‌θ−gt)2
=√u2cos2θ+u2sin2θ+g2t2−2‌ugt‌sin‌θ
=√u2(cos2θ+sin2θ)+g2t2−2‌ugt‌sin‌θ
∴ KE=

1

2

m(u2+g2t2−2‌ugt‌sin‌θ)
i.e., KE∝t2
Hence, graph will be parabolic intercept on y-axis.
Hence, the graph between KE and t.
Now, in case of height
KE=

1

2

m(v2)‌and‌v2=(u2−2‌gy)
∴ KE=

1

2

m(u2−2‌gy)
KE=−mgy+

1

2

mu2
Intercept on y-axis =

1

2

mu2

Now, KE=

1

2

mv2