Explore Westonci.ca, the top Q&A platform where your questions are answered by professionals and enthusiasts alike. Explore thousands of questions and answers from a knowledgeable community of experts on our user-friendly platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
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
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.