A 0.5kg basketball moving at 1.4m/s collides with a 1.2kg bowling ball moving at 3.5m/s. After collision, the velocity of the basketball was increased by 0.4m/s. What is the final velocity of the bowling ball?
Solution.
m1=0.5kg;m_1=0.5 kg;m1=0.5kg;
v1=1.4m/s;v_1=1.4 m/s;v1=1.4m/s;
m2=1.2kg;m_2=1.2 kg;m2=1.2kg;
v2=3.5m/s;v_2=3.5 m/s;v2=3.5m/s;
u1=1.8m/s;u_1=1.8 m/s;u1=1.8m/s;
u2−?;u_2 -?;u2−?;
m1v1−m2v2=−m2u2−m1u1;m_1v_1-m_2v_2=-m_2u_2-m_1u_1;m1v1−m2v2=−m2u2−m1u1;
m2u2=−m1v1+m2v2−m1u1;m_2u_2=-m_1v_1+m_2v_2-m_1u_1;m2u2=−m1v1+m2v2−m1u1;
u2=−m1v1+m2v2−m1u1m2;u_2=\dfrac{-m_1v_1+m_2v_2-m_1u_1}{m_2};u2=m2−m1v1+m2v2−m1u1;
u2=−0.5⋅1.4+1.2⋅3.5−0.5⋅1.81.2=2.2m/s;u_2=\dfrac{-0.5\sdot1.4+1.2\sdot3.5-0.5\sdot1.8}{1.2}=2.2 m/s;u2=1.2−0.5⋅1.4+1.2⋅3.5−0.5⋅1.8=2.2m/s;
Answer: u2=2.2m/s.u_2=2.2 m/s.u2=2.2m/s.
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