Sagot :
Its about momentum. Momentum (p)=mass(m)xvelocity(v)
So for the first ball P=4x8=32kgm/s
For the second the momentum is zero as it is still.
So overall momentum its 32kgm/s
Momentum has to be conserved
After the collision the momentum of the 4kg ball is 4x4.8=19.2kgm/s
As momentum is conserved 32-19.2=12.8kgm/s remaining
So rearrange for velocity so v=p/m=12.8/1=12.8m/s for the 1kg ball
So for the first ball P=4x8=32kgm/s
For the second the momentum is zero as it is still.
So overall momentum its 32kgm/s
Momentum has to be conserved
After the collision the momentum of the 4kg ball is 4x4.8=19.2kgm/s
As momentum is conserved 32-19.2=12.8kgm/s remaining
So rearrange for velocity so v=p/m=12.8/1=12.8m/s for the 1kg ball
Answer: The velocity of ball having less mass is 12.8 m/s
Explanation:
To calculate the velocity of the ball having less mass after the collision, we use the equation of law of conservation of momentum, which is:
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]
where,
[tex]m_1[/tex] = mass of ball 1 = 4.0 kg
[tex]u_1[/tex] = Initial velocity of ball 1 = 8.0 m/s
[tex]v_1[/tex] = Final velocity of ball 1 = 4.8 m/s
[tex]m_2[/tex] = mass of ball 2 = 1.0 kg
[tex]u_2[/tex] = Initial velocity of ball 2 = 0 m/s
[tex]v_2[/tex] = Final velocity of ball 2 = ?
Putting values in above equation, we get:
[tex](4.0\times 8.0)+(1.0\times 0)=(4.0\times 4.8)+(1.0\times v_2)\\\\v_2=\frac{32-19.2}{1}=12.8m/s[/tex]
Hence, the velocity of ball having less mass is 12.8 m/s