Answer:
a) [tex]\mathbf{|\Delta p^{\to}_{11}| = 0 \ kg.m/s}[/tex]
b) [tex]\mathbf{|\Delta p^{\to}_{12}| = 500 \ kg.m/s}[/tex]
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:
[tex]v_1^{\to} = v \hat _v_1} \\ \\ v_1^{\to} = (5)(0,1,0)\\ \\ v_1^{\to} = (0,5,0) \ m/s[/tex]
the momentum will be:
[tex]p_1^{\to} = m v_1^{\to} \\ \\ p_1^{\to} = (50)(0,5,0) \\ \\ p_1^{\to} = (0,250,0) \ kg.m/s[/tex]
the change in momentum now is [tex]\Delta p = p_1^{\to} -p_1^{\to}[/tex] since that is the child's momentum initially.
∴ [tex]\Delta p =(0,250,0) - (0,250,0)[/tex]
[tex]\mathbf{\Delta p =(0,0,0) \ kg.m/s}[/tex]
By subtracting the two vector graphically as being asked in the question, we have :
[tex]|\Delta p^{\to}_{11}| = \sqrt{(0)^2+(0)^2 +(0)^2 }[/tex]
[tex]\mathbf{|\Delta p^{\to}_{11}| = 0 \ kg.m/s}[/tex]
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:
[tex]v_2^{\to} = v \hat _v_2} \\ \\ v_2^{\to} = (5)(0,-1,0)\\ \\ v_2^{\to} = (0,-5,0) \ m/s[/tex]
the momentum will be:
[tex]p_2^{\to} = m v_2^{\to} \\ \\ p_2^{\to} = (50)(0,-5,0) \\ \\ p_2^{\to} = (0,-250,0) \ kg.m/s[/tex]
the change in momentum now is [tex]\Delta p = p_2^{\to} -p_1^{\to}[/tex] since that is the child's momentum initially.
∴ [tex]\Delta p =(0,-250,0) - (0,250,0)[/tex]
[tex]{\Delta p =(0,-500,0) \ kg.m/s[/tex]
By subtracting the two vector graphically as being asked in the question, we have :
[tex]|\Delta p^{\to}_{12}| = \sqrt{(0)^2+(-500)^2 +(0)^2 }[/tex]
[tex]|\Delta p^{\to}_{12}| = \sqrt{250000}[/tex]
[tex]\mathbf{|\Delta p^{\to}_{12}| = 500 \ kg.m/s}[/tex]
