Question

5kg ball a rolls on smooth surface with a velocity 10m/s collides with nother ball b of mass 10kg moving opposite to ball a with a velocity 4m/s the coefficient restitution is 0.6 between the balls .Calculate the velocities of balls immediately after impact

Answer 1

Explanation:

please conserve momentum and solve see pic posted above

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Answer 2

Velocities of ball a -4.94 m/s and ball b 3.46 m/s

Explanation:

given data

Mass of ball a, m1= 5 kg

Velocity of ball b, u1  = 10 m/s

Mass of ball a, m2 = 10 kg

Velocity of ball b,  u2  = 4 m/s

Coefficient restitution c = 0.6

solution

we use here equation of coefficient restitution  to get velocity

$v{2}-v{1}=c \times (u{1}+u{2})$ ..................1

put here value we get

$v{2}-v{1}= 0.6 \times (10+4)$

$v{2}-v{1}= 8.4$    ..........2

and

now we use here conservation of momentum to get velocity

$m{1}\times u{1}-m{2}\times u{2}=m{1}\times v{1}\times +m{2}\times v{2}$   .......................3

put here value

$5\times 10-10\times 4=5 \times v{1}+10\times v{2}$  

$v{1}+2\times v{2}=2$     ...........4

now we equate equation From equation 2 and 4

$3\times v{2}=10.4$

$v{2}=\dfrac{10.4}{3}$

$v{2}=3.46$  m/s

and

from equation 2

$3.46-v{1}=8.4$

$v{1}=-4.94$

so that velocities of ball a -4.94 m/s and ball b 3.46 m/s

learn more :

1. https://brainly.in/question/5223908

2. https://brainly.in/question/14699703