Respuesta :
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
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
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.
[tex]m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}[/tex].
Here, [tex]m_{1}[/tex] is the mass of larger marble, [tex]m_{2}[/tex] is the mass of smaller marble, [tex]u_{1}[/tex] is the velocity of the larger marble and [tex]u_{2}[/tex] is the velocity of smaller marble before collision and [tex]v_{1}[/tex] is the velocity of the larger marble and [tex]v_{2}[/tex] is the velocity of smaller marble after collision.
Given, [tex]m_{1} = 2g[/tex],[tex]m_{2} = 1.25\ g[/tex],[tex]u_{1} =0\ m /s[/tex],[tex]u_{2} =8\ m/s[/tex], [tex]v_{1} =?[/tex] and [tex]v_{2} =1\ m/s[/tex].
Substituting these values in above equation,
[tex]2g\times 0 + 1.25g\times 8\ m/s = 2g\times v_{1} + 1.25g\times\ 1 m/s\\\\ v_{1}=\frac{10-1.25}{2} =4.375\simeq 4m/s.[/tex]
Thus, the velocity of 2 g marble after the collision is [tex]4m/s[/tex]
