Question

if the linear momentum of the object is increased by 0.1% then the kinetic energy is increased by
​

Answer 1

Explanation:

Answer:

44

%

Explanation:

The applicable kinematic expression is

v

2

−

u

2

=

2

a

s

where

v

is final velocity,

u

is initial velocity,

a

is acceleration and

s

is displacement.

Since the car needs to stop, hence in the first instance distance traveled

s

1

, when

r

is the retardation

0

2

−

u

2

1

=

−

2

r

s

1

⇒

s

1

=

u

2

1

2

r

in the second instance given

u

2

=

1.2

u

1

Hence, distance traveled with same deceleration

⇒

s

2

=

(

1.2

u

1

)

2

2

r

Fractional increase in stopping distance

s

2

−

s

1

s

1

=

(

1.2

u

1

)

2

2

r

−

u

2

1

2

r

u

2

1

2

r

⇒

s

2

−

s

1

s

1

=

1.44

−

1

⇒

s

2

−

s

1

s

1

=

0.44

Percent increase in stopping distance

=

0.44

×

100

=

44