Question

A body of mass 2 kg makes an elastic collision with a second body at rest and continues to move in the original direction but with one fourth of its original speed. What is the mass of the second body?

Answer 1

Answer:

By conservation of linear momentum

\quad 2{ v }_{ 0 }=2\left( \cfrac { { v }_{ 0 } }{ 4 } \right) +mv\Rightarrow 2{ v }_{ 0 }=\cfrac { { v }_{ 0 } }{ 2 } +mv\Rightarrow \cfrac { { 3v }_{ 0 } }{ 2 } =mv....(1)\quad2v

0

=2(

4

v

0

)+mv⇒2v

0

=

2

v

0

+mv⇒

2

3v

0

=mv....(1)

since collision is elastic

{ v }_{ separation }={ v }_{ apprach }\Rightarrow v-\cfrac { { v }_{ 0 } }{ 4 } ={ v }_{ 0 }\Rightarrow \cfrac { { 5v }_{ 0 } }{ 4 } =v....(2)v

separation

=v

apprach

⇒v−

4

v

0

=v

0

⇒

4

5v

0

=v....(2)

equating (2) and (1)

\cfrac { { 3v }_{ 0 } }{ 4 } =m\left( \cfrac { { 5v }_{ 0 } }{ 4 } \right) \Rightarrow m=\cfrac { 6 }{ 5 } =1.2kg

4

3v

0

=m(

4

5v

0

)⇒m=

5

6

=1.2kg