Question

An experiment is performed aboard the International Space Station to verify that linear momentum is conserved during collisions in a zero-g environment. The experiment involves a 3-D completely inelastic collision of three drops of honey. At the moment just before they all collide, the masses and velocities of the drops are

m1 = 31.5 g V1 = (12.7 m/s)x
m2 = 49.9 g V2 = (12.5 m/s)y
m3 = 78.1 g V3 = (15.9 m/s)z
What is your prediction for the speed, V, of the combined honey drop after the collision?
V = _______________________ m/s

Answer 1

The speed of the combined honey drop after the collision is 9.05 m/s.

Step-by-step explanation:

Given that,

The masses and velocities of the drops are

`m_(1)=31.5\ g`

`v_(1)=12.7\ m/s`

`m_(2)=49.9\ g`

`v_(1)=12.5\ m/s`

`m_(1)=78.1\ g`

`v_(1)=15.9\ m/s`

We need to calculate the total mass

Using formula of masses

`m=m_(1)+m_(2)+m_(3)`

Put the value into the formula

`m=31.5+49.9+78.1`

`m=159.5\ g`

We need to calculate the velocity in x- direction

Using conservation of momentum

`m_(1)v_(1)=(m_(1)+m_(2)+m_(3))v_(x)`

`v_(x)=(m_(1)v_(1))/(m_(1)+m_(2)+m_(3))`

Put the value into the formula

`v_(x)=(31.5*12.7)/(159.5)`

`v_(x)=2.50\ m/s`

We need to calculate the velocity in y- direction

Using conservation of momentum

`m_(2)v_(2)=(m_(1)+m_(2)+m_(3))v_(y)`

`v_(y)=(m_(2)v_(2))/(m_(1)+m_(2)+m_(3))`

Put the value into the formula

`v_(y)=(49.9*12.5)/(159.5)`

`v_(y)=3.91\ m/s`

We need to calculate the velocity in z- direction

Using conservation of momentum

`m_(3)v_(3)=(m_(1)+m_(2)+m_(3))v_(z)`

`v_(z)=(m_(3)v_(3))/(m_(1)+m_(2)+m_(3))`

Put the value into the formula

`v_(z)=(78.1*15.9)/(159.5)`

`v_(z)=7.78\ m/s`

We need to calculate the combined honey drop after the collision

Using formula of velocity

`v=v_(x)+v_(y)+v_(z)`

Put the value into the formula

`v=2.50i+3.91j+7.78k`

The magnitude of velocity

`v=√((2.50)^2+(3.9)^2+(7.78)^2)`

`v=9.05\ m/s`

Hence, The speed of the combined honey drop after the collision is 9.05 m/s.