Respuesta :
Answer:
20 m/s
Explanation:
mass of puck 1, m1 = 0.5 kg
initial velocity, u1 = 40 m/s
mass of puck 2, m2 = 1 kg
initial velocity of puck 2, u2 = 0 m/s
final velocity of puck 1, v1 = 0 m/s
Let the final velocity of puck 2 is v2.
As there is not external force acting on the system, the momentum of the system is conserved.
m1 x u1 + m2 x u2 = m1 x v1 + m2 x v2
0.5 x 40 + 1 x 0 = 0.5 x 0 + 1 x v2
20 = v2
v2 = 20 m/s
Thus, the velocity of puck 2 after the collision is 20 m/s.
Answer:
The final velocity of puck 2 after the collision is 20 m/s .
Explanation:
Let us assume , right direction is positive and left direction is negative .
Given :
Mass of puck 1 and 2 of 0.5 kg and 1 kg respectively .
Initial velocity of puck 1 is 40 m/s .
Initial velocity of puck 2 is 0 m/s .
Final velocity of puck 1 is 0 m/s .
Let , us assume final velocity of puck 2 is [tex]v_2[/tex] .
Therefore , by conservation of momentum :
[tex]0.5\times 40+1\times 0=0.5\times 0+1\times v_2\\v_2=20\ m/s[/tex]
Therefore , the final velocity (in m/s) of puck 2 after the collision is 20 m/s .
