EAMCET Engineering 2009 Solved Paper

Momentum, ‌‌p=m⋅v‌⇒‌‌v=(‌

p

m

)
‌‌ Kinetic energy, ‌KE=‌

1

2

mv2
‌=‌

1

2

m(‌

p2

m2

)=‌

1

2m

p2
‌⇒‌‌KE∝p2‌‌(∵‌

1

2m

=‌ constant ‌)
Hence, the graph between KE and p2 will be linear as shown below
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)
‌⇒KE∝t2
Hence, graph will be parabolic with intercept on y -axis.
Hence, the graph between KE and tNow, in case of height
KE=‌

1

2

m(v2)
and
‌∴‌‌KE=‌

1

2

m(u2−2gy)
‌KE=−mgy+‌

1

2

mu2
‌‌ Intercept on ‌y‌-axis ‌=‌

1

2

mu2Now,
‌KE=‌

1

2

mv2
‌KE=‌

1

2

m(‌

x

t

)2KE∝x2 . Thus graph between KE and x will be parabolic.