Answer:
a) Final velocity of second bowling pin is 2.5m/s.
b) Final velocity of second bowling pin is 3 m/s.
Explanation:
Let 'm' be the mass of both the bowling pin -
m = 1.5 kg
Initial velocity of first bowling pin -
[tex]v_{1} = 3 m/s[/tex]
In any type of collision between two bodies in horizontal plane , momentum is conserved along the line of impact.
a) Since , initial velocity of second bowling pin is 0 m/s -
Initial momentum ,
[tex]p_{1} = mv_{1}[/tex]
Final velocity of first bowling pin , [tex]v_{2} = 0.5m/s[/tex] [Considering initial direction of motion of the first bowling pin to be positive]
Let [tex]u_{2}[/tex] be the final velocity of the second bowling pin.
∴ Final momentum ,
[tex]p_{2} = mv_{2} + mu_{2}[/tex].
Now ,
[tex]p_{1} = p_{2}[/tex]
∴[tex]mv_{1} = mv_{2} + mu_{2}[/tex]
∴[tex]u_{2}[/tex] = 3 - 0.5 = 2.5 m/s
∴ Final velocity of second bowling pin is 2.5 m/s.
b) Since , initial velocity of second bowling pin is 0 m/s -
Initial momentum ,
[tex]p_{1} = mv_{1}[/tex]
Final velocity of first bowling pin , [tex]v_{2} = 0m/s[/tex] [given][Considering initial direction of motion of the first bowling pin to be positive]
Let [tex]u_{2}[/tex] be the final velocity of the second bowling pin.
∴ Final momentum ,
[tex]p_{2} = mv_{2} + mu_{2}[/tex].
Now ,
[tex]p_{1} = p_{2}[/tex]
∴[tex]mv_{1} = mv_{2} + mu_{2}[/tex]
∴[tex]u_{2}[/tex] = 3 - 0 = 3 m/s
∴ Final velocity of second bowling pin is 3 m/s.
