Question

During a game of marbles, a player is trying to win a 2 g marble that is at rest. the player flicks a 1.25 g marble at a velocity of 8 at the larger marble that is sitting still. after the collision, the smaller marble has a velocity of 1 . what is the velocity of the 2 g marble after the collision? round your answer to the nearest whole number

Answer 1

To determine the velocity of 2 g marble after the collision, we use the equation for conservation of momentum.

Momentum before the collision= Momentum after the collision.

`m_(1) u_(1) + m_(2) u_(2) = m_(1) v_(1) + m_(2) v_(2)`.

Here,
`m_(1)` is the mass of larger marble,
`m_(2)` is the mass of smaller marble,
`u_(1)` is the velocity of the larger marble and
`u_(2)` is the velocity of smaller marble before collision and
`v_(1)` is the velocity of the larger marble and
`v_(2)` is the velocity of smaller marble after collision.

Given,
`m_(1) = 2g`,
`m_(2) = 1.25\ g`,
`u_(1) =0\ m /s`,
`u_(2) =8\ m/s`,
`v_(1) =?` and
`v_(2) =1\ m/s`.

Substituting these values in above equation,

`2g* 0 + 1.25g* 8\ m/s = 2g* v_(1)  + 1.25g*\ 1 m/s\\\\ v_(1)=(10-1.25)/(2) =4.375\simeq 4m/s.`

Thus, the velocity of 2 g marble after the collision is
`4m/s`

Answer 2

momentum before = momentum after
0.00125kg*8+0.002*0=0.00125*1+0.002*x
0.01=0.00125+0.002x
0.01-0.00125=0.002x
0.00875=0.002x
0.00875/0.002=x
4=x
the velocity of the larger marble is 4