Question

Two snowmobiles, Frosty and Snowflake, of equal masses suffer a rear-end collision. Just Before the collision, Frosty's velocity is 25 m/s east and Snowflake's velocity is 10 m/s in the same firection. Just after the collision, Frosty is moving 13m/s towards the east.

A. What is Snow flakes velocity immediately after the collision?


B. Was this an elastic collision? Justify your answer by calculation.


Please. I am sooo lost..

Answer 1

A. 22
`ms^(-1)`

B. NO

Step-by-step explanation:

A.

During a collision , the net external force on the system is zero .

hence , the total momentum of the system can be CONSERVED .

let the mass of frosty and snowflake be m ;

• initial velocity of frosty be
  `v_{F_(i) }` = 25
  `ms^(-1)`
• final velocity of frosty be
  `v_{F_(f) }` = 13
  `ms^(-1)`

• initial velocity of snowflake be
  `v_{S_(i) }` = 10
  `ms^(-1)`

• final velocity of snowflake be
  `v_{S_(f) }` = x
  `ms^(-1)`

therefore from principle of conservation of momentum ,

`m*v_{F_(i) } +  m*v_{S_(i) } = m*v_{F_(f) } +  m*v_{S_(f) }`

so ,

`m*25 + m*10 =m*13 +m*x`

`x = 25+10-13 = 22 ms^(-1)`

answer for A. 22
`ms^(-1)`

B.

The collision is NOT ELASTIC.

this is because if it had been elastic , the coefficient of restitution should have been 1 but it isnot i.e

`e =\frac{ v_{F_(f) } - v_{S_(f) } }{ v_{S_(i) } - v_{F_(i) } } \\=(13-22)/(10-25) \\=(9)/(15) = (3)/(5) =<strong> 0.6` ≠ 1

thus it's an inelastic collision.