Question

A body having mass m1 travelling  with velocity  u  makes an head on elastic collision with the stationary body  B of mass m2.....

(a) Calculate the final velocities of A and B
(b) Calculate the Final Kinetic energy of both the mass
(c) Calculate the ratio of the kinetic energy transferred to m2 to the original kinetic energy
(d) For what value of m2, all the energy is transferred to the stationary object
(e) Calculate the velocity of the Center of mass before collision and after collision.


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Answer 1

Explanation:

1.Let u¹ be the initial velocity of mass m ¹and v ¹be the final velocity of the mass m 1.

As it perfect elastic collision , from law of conservation of momentum, we get

m¹ u ¹ =m ¹ v ¹ +m ² v ²

m ² v ² =m¹ u ¹−m ¹ v ¹

m ² v ² =m¹ (u ¹−v ¹ )----------(1)

Again from law of conservation of kinetic energy we get,

½ m¹u ²/¹= ½ m ¹v ²/¹ + ½ m ²v ²/²

m ¹ u ½=m ¹ v½+m ²v²/²

-------------(2)

Dividing 2 and 1 we get,

v 2=u 1 +v 1

-------(3)

Putting 3 in 1 we get,

m 2m1u1−m 1v/m1 =u 1+v 1

u 1 −m v 1 =m 2 uu1+m 2

v 1

u

1

(m

1

−m

2

)=v

1

(m

2

+m

1

)

Thus,

K.E

i

K.E

i

−K.E

f

=1−

(1/2)m

1

u

1

2

(1/2)m

1

v

1

2

=1−

u

1

(v

1

)

2

1−[

m

1

+m

2

m

1

−m

2

]

2

=

[m

1

+m

2

]

2

4m

1

m

2

Answer 2

Let masses be m, km Let

initial velocity be v. vel of m

after collision be

$\frac{v}{1.5} = \frac{2v}{3}$

conservation of momentum mv =$\frac{2mv}{3} + {kmv}^{1}$

${v}^{1} = \frac{v}{3k}$

For elastic collision, e = 1

$v = \frac{v}{3k} - \frac{2v}{3}$

$k = \frac{1}{5}$

Ratio is 1:1/5 i.e., 5:1