Question

A 50 kg child is riding on a carousel (merry-go-round) at a constant speed of 5 m/s. What is the magnitude of the change in the child's momentum ∣Δp⃗ ∣∣ in going all the way around (360∘)?In going halfway around (180∘)? It is very helpful to draw a diagram, and to do the vector subtraction graphically.

Answer 1

a)
`\mathbf\Delta p^(\to)_(11)`

b)
`\mathbf`

Step-by-step explanation:

From the image attached below.

Suppose the child goes all the way around, i.e., 360, the child will execute a movement of 1 complete revolution and be at his starting point. At that point, the velocity vector is towards the y-direction.

Thus, the velocity of the child is:

`v_1^(\to) = v \hat _v_1} \\ \\ v_1^(\to) = (5)(0,1,0)\\ \\ v_1^(\to) = (0,5,0) \ m/s`

the momentum will be:

`p_1^(\to) = m v_1^(\to) \\ \\ p_1^(\to) = (50)(0,5,0) \\ \\ p_1^(\to) = (0,250,0) \ kg.m/s`

the change in momentum now is
`\Delta p = p_1^(\to) -p_1^(\to)` since that is the child's momentum initially.

∴
`\Delta p =(0,250,0) - (0,250,0)`

`\mathbf{\Delta p =(0,0,0) \ kg.m/s}`

By subtracting the two vector graphically as being asked in the question, we have :

`|\Delta p^(\to)_(11)| = √((0)^2+(0)^2 +(0)^2 )`

`\mathbf`

b) In going halfway around (180°), the child will be opposite with respect to the starting point. Hence, the velocity vector will be in the negative y-direction.

Thus, the velocity of the child is:

`v_2^(\to) = v \hat _v_2} \\ \\ v_2^(\to) = (5)(0,-1,0)\\ \\ v_2^(\to) = (0,-5,0) \ m/s`

the momentum will be:

`p_2^(\to) = m v_2^(\to) \\ \\ p_2^(\to) = (50)(0,-5,0) \\ \\ p_2^(\to) = (0,-250,0) \ kg.m/s`

the change in momentum now is
`\Delta p = p_2^(\to) -p_1^(\to)` since that is the child's momentum initially.

∴
`\Delta p =(0,-250,0) - (0,250,0)`

`{\Delta p =(0,-500,0) \ kg.m/s`

By subtracting the two vector graphically as being asked in the question, we have :

`|\Delta p^(\to)_(12)| = √((0)^2+(-500)^2 +(0)^2 )`

`|\Delta p^(\to)_(12)| = √(250000)`

`\mathbf\Delta p^(\to)_(12)`