# Answer to Question #71307 in Mechanics | Relativity for salim

Question #71307

The mass of the blue puck shown below is 15.0% greater than the mass of the green puck. Before colliding, the pucks approach each other with momenta of equal magnitudes and opposite directions, and the green puck has an initial speed of 13.0 m/s. Find the speeds the pucks have after the collision if half the kinetic energy of the system becomes internal energy during the collision.

Expert's answer

M - mass of the blue puck

m - mass of the green puck

v1 – speed of the blue puck before the collision

v2=13.0m/s - speed of the green puck before the collision

v’1 – speed of the blue puck after the collision

v’2 - speed of the green puck after the collision

According to the task:

M=m+0.15*m=> M/m=1.15

Mv_1=mv_(2 )=> v_1=(mv_(2 ))/M

After the collision:

.█(1/2 ((Mv_1^2)/2+(mv_2^2)/2)=(Mv_1^'2)/2+(mv_2^'2)/2 -The law of energy conservation @Mv_1^'=mv_2^'-The law of momenta conservation)

Find v’1 and v’2 from this system of equations:

v_2^'=v_2/√2≈9.19 m/s

v_1^'=m/M v_2/√2≈7.99 m/s

Answer:

v_2^'≈9.19 m/s

v_1^'≈7.99 m/s

m - mass of the green puck

v1 – speed of the blue puck before the collision

v2=13.0m/s - speed of the green puck before the collision

v’1 – speed of the blue puck after the collision

v’2 - speed of the green puck after the collision

According to the task:

M=m+0.15*m=> M/m=1.15

Mv_1=mv_(2 )=> v_1=(mv_(2 ))/M

After the collision:

.█(1/2 ((Mv_1^2)/2+(mv_2^2)/2)=(Mv_1^'2)/2+(mv_2^'2)/2 -The law of energy conservation @Mv_1^'=mv_2^'-The law of momenta conservation)

Find v’1 and v’2 from this system of equations:

v_2^'=v_2/√2≈9.19 m/s

v_1^'=m/M v_2/√2≈7.99 m/s

Answer:

v_2^'≈9.19 m/s

v_1^'≈7.99 m/s

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