A blue puck with mass 0.0400 kg, sliding with a speed of 0.200 m/s on a frictionless, horizontal surface, makes a perfectly elastic head‐on collision with a red puck with mass m, initially at rest. After the collision, the velocity of the blue puck is 0.0500 m/s in the same direction as its initial velocity. (a) Find the mass m of the red puck, and (b)the velocity (magnitude and direction) of the red puck after the collision

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WinterWhitney WinterWhitney · Answer 1 · 2020-04-12T08:28:47+02:00

Answer:

Part(a): The velocity of the red puck is [tex]\bf{0.2500~m/s}[/tex] and the direction will be the same as initial direction of blue puck.

Part(b): The mass of the red puck is [tex]\bf{0.024~Kg}[/tex].

Explanation:

Given:

The mass of the blue puck, [tex]m_{b} = 0.0400~Kg[/tex]

The initial velocity of the blue puck, [tex]v_{ib} = 0.200~m/s[/tex]

The final velocity of the blue puck, [tex]v_{fb} = 0.0500~m/s[/tex]

The initial velocity of the red puck, [tex]v_{ir} = 0~m/s[/tex]

Consider the mass of the red puck be [tex]m_{r}[/tex] and its final velocity be [tex]v_{fr}[/tex].

Part(a):

The relation between the relative velocities of the particles is given by

[tex]v_{ib} - v_{ir} = v_{fr} - v_{fb}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)[/tex]

Substitute the values in equation (1)

[tex]~~~~&& 0.200~m/s - 0 = v_{fr} - 0.0500~m/s\\&or,& v_{fr} = 0.2500~m/s[/tex]

Part(b):

From momentum conservation,

[tex]m_{b}v_{ib} + m_{r}v_{ir} = m_{b}v_{fb} + m_{r}v_{fr}~~~~~~~~~~~~~~~~~~~~~~~~(2)[/tex]

Substitute the values in equation (2).

[tex]~~~~(0.0400~Kg)(0.200~m/s) - 0 = m_{r}(0.2500~m/s) + (0.0400~Kg)(0.05~m/s)\\&or,& m_{r} = 0.024~Kg[/tex]