Answer:
[tex]v_{f2}[/tex] =6.5%[tex]v_{i1}[/tex]
Explanation:
Mass of the ball: [tex]m_{1} =0.12kg[/tex]]
Initial velocity of the ball: [tex]v_{i1}[/tex]
final velocity of the ball: [tex]v_{f1}[/tex] which is -30/100 of [tex]v_{i1}[/tex] =[tex]-0.3v_{i1}[/tex]
Mass of the bottle: [tex]m_{2} =2.4kg[/tex]
Initial velocity of the bottle: [tex]v_{i2}=0m/s[/tex]
final velocity of the bottle: [tex]v_{f2}[/tex] is unknown (to find)
by using conservation momentum, which stated that the initial momentum is equal to the final momentum.
[tex]m_{1} v_{i1} +m_{2} v_{i2} =m_{1} v_{f1} +m_{2} v_{f2}[/tex]
so since the bottle is at rest firstly, therefore [tex]v_{i2} =0[/tex]
[tex]m_{1} v_{i1} +m_{2} (0) =m_{1} v_{f1} +m_{2} v_{f2}[/tex]
[tex]m_{1} v_{i1} =m_{1} v_{f1} +m_{2} v_{f2}[/tex] equation 1
so now substitute [tex]v_{f1}[/tex] into equation 1
[tex]m_{1} v_{i1} =m_{1} (-0.3v_{i1} ) +m_{2} v_{f2}[/tex]
[tex]m_{1} v_{i1} = -0.3m_{1}v_{i1} +m_{2} v_{f2}[/tex]
collect the like terms
[tex]m_{1} v_{i1} +0.3m_{1}v_{i1} =m_{2} v_{f2}[/tex]
[tex]1.3m_{1} v_{i1} =m_{2} v_{f2}[/tex]
divide both side by [tex]m_{2}[/tex]
[tex]v_{f2}=\frac{1.3m_{1} v_{i1}}{m_{2} }[/tex]
Now substitute
[tex]v_{f2} =\frac{1.3*0.12*v_{i1}}{2.4}\\v_{f2} =\frac{0.156v_{i1} }{2.4} \\v_{f2} =0.065v_{i1}[/tex]
[tex]v_{f2} =[/tex]6.5%[tex]v_{i1}[/tex]
Answer:
v2 = 6.5% of the ball's initial velocity
Explanation:
Let the mass of the ball and bottle be m1 and m2 respectively.
m1 = 0.12kg , m2 = 2.4kg
Let the initial and final velocity of ball be u1 and
v1 = -0.3u1 while that of the bottle be u2 and v2
The bottle was initially at rest so u2 = 0m/s
From the principle of conservation of momentum, momentum before collision is equal to momentum after collision.
m1u1 + m2u2 = m1v1 + m2v2
0.12u1 + m2×0 = 0.12(–0.3u1) + 2.4v2
0.12u1 = –0.036u1 + 2.4v2
Collecting like terms
2.4v2 = 0.12u1 + 0.036u1
2.4v2 = 0.156u1
v2 = 0.156u1/2.4
v2 = 0.065u1 = 6.5%u1
