Asked by Saiffah
A ball moving with a speed of 9 m/s strikes an identical stationary ball such that after the collision the dierection of each ball makes an angle of 30° with the original line of motion. Find the speeds of the two balls after the collision.
Answers
Answered by
Ashwin Chellappan
u1 =9m/s,u2 =0
Initial momentum = Final momentum along original line of motion Mx9+mx0=mv1cos30+mv2 cos30
v1 +v2 =6√3m/s
Direction perpendicular to the original line
0 = mv1sin30 – mv2sin30
v1 = v2
v1 =v2 =3√3m/s
Total K.E before collision = 1⁄2 x mx 92
= 40.5 m
After collision = 1⁄2 mx (3√3)2 + 1⁄2 x m x (3√3)2
= 27 m K.E is not conserved.
Initial momentum = Final momentum along original line of motion Mx9+mx0=mv1cos30+mv2 cos30
v1 +v2 =6√3m/s
Direction perpendicular to the original line
0 = mv1sin30 – mv2sin30
v1 = v2
v1 =v2 =3√3m/s
Total K.E before collision = 1⁄2 x mx 92
= 40.5 m
After collision = 1⁄2 mx (3√3)2 + 1⁄2 x m x (3√3)2
= 27 m K.E is not conserved.
Answered by
Anupama Joy
u1=9m/s, u2=0
along the vertical,
mv1sin30-mv2sin30=0
v1=v2
along the horizontal,
mv1cos30+mv2cos30=
along the vertical,
mv1sin30-mv2sin30=0
v1=v2
along the horizontal,
mv1cos30+mv2cos30=
Answered by
Sahaj
Thnx a lot for ur help 😊......
Answered by
🤡🤡🤡🤡🤡🤡
Nothing helped to me as the question asked i.e, is it elastic oe inelastic but no on has answered this question so just stupid
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