Question #195259

A sphere of mass 12kg and another one of mass 8 kg are moving towards each other in a smooth linear groove. The speed of the heavier sphere is 3 times that of the lighter sphere . After collision the heavier sphere has a speed of 4/1/6 m/s and the lighter sphere a speed of 6m/s . Calculate the speeds of the sphere before collisions


1
Expert's answer
2021-05-19T17:03:54-0400

Solution.

m1=12kg;m_1=12kg;

m2=8kg;m_2=8kg;

v01=3v;v_{01}=3v;

v02=v;v_{02}=v;

v1=416m/s;v_1=4\dfrac{1}{6}m/s;

v2=6m/s;v_2=6m/s;

m1v01+m2v02=m1v1+m2v2;m_1\overrightarrow{v_{01}}+m_2\overrightarrow{v_{02}}=m_1\overrightarrow{v_{1}}+m_2\overrightarrow{v_{2}};

m1v01m2v02=m1v1+m2v2;m_1v_{01}-m_2v_{02}=m_1v_1+m_2v_2;

123v8v=12416+86;12\sdot3v-8v=12\sdot4\dfrac{1}{6}+8\sdot6;

36v8v=50+48;36v-8v=50+48;

26v=98;26v=98;

v=98/26;v=98/26;

v=31013;v=3\dfrac{10}{13};

v01=331013=11413m/s;v_{01}=3\sdot3\dfrac{10}{13}=11\dfrac{4}{13}m/s;

v02=31013m/s;v_{02}=3\dfrac{10}{13}m/s;

Answer: v01=11413m/s;v02=31013m/s.v_{01}=11\dfrac{4}{13}m/s;v_{02}=3\dfrac{10}{13}m/s.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS