Question

Two 2.0 kg bodies, A and B, collide. The velocities before the collision are

v→A=(15i^+30j^) m/s
and
v→B=(−10i^+5.0j^) m/s
. After the collision,
v→A=(−5.0i^+20j^) m/s
. What are (a) the final velocity of B and (b) the change in the total kinetic energy (including sign)?

Answer 1

Step-by-step explanation:

Given

mass of body A
`m_a=2 kg`

mass of body
`m_b=2 kg`

Velocity before Collision is
`u_a=15\hat{i}+30\hat{j}`

`u_b=-10\hat{i}+5\hat{j} m/s`

after collision
`v_a=-5\hat{i}+20\hat{j} m/s`

let
`v_b_x` and
`v_b_y` velocity of B after collision in x and y direction

conserving momentum in x direction

`m_a* 15+m_b* (-10)=m_a* (-5)+m_b* (v_b_x)`

as
`m_a=m_b` thus

`15-10=-5+v_b_x`

`v_b_x=10 m/s`

Conserving momentum in Y direction

`m_a* 30+m_b* 5=m_a* 20+m_b* (v_b_y)`

`30+5=20+v_b_y`

`v_b_y=15 m/s`

thus velocity of B after collision is

`v_b=10\hat{i}+15\hat{j}`

(b)Change in total Kinetic Energy

Initial Kinetic Energy of A And B

`K.E._a=(1)/(2)* 2(√(15^2+30^2))^2=1125 J`

`K.E._b=(1)/(2)* 2(√(10^2+5^2))^2=125 J`

Total initial Kinetic Energy =1250 J

Final Kinetic Energy of A And B

`K.E._a=(1)/(2)* 2(√(5^2+20^2))^2=425 J`

`K.E._b=(1)/(2)* 2(√(10^2+15^2))^2=325 J`

Final Kinetic Energy
`=425+325=750 J`

Change
`\Delta K.E.=1250-750=500 J`