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Grade 11Mechanics

A 0.5 kg ball moving with velocity 12m/s strikes a hard wall at an angle 30 degree with the wall. It is reflected with the same speed at the same angle.If the ball is in cantact with the wall for 0.25s ,the average force on the wall is

sabita

10 Years agoGrade 11

4 Answers

Amogh Gajare

10 Years ago

F (average) = m (vf - vi)/t

V_f=0 m/s

V_i=12 m/s

Therefore by the formula,

F_avg.= m (vf - vi)/t

= 0.5 (0-12)/0.25

= -24 N

Tell if any problem.

Sneha suresh

8 Years ago

Let ,vector OA be the momentum before collision and vector OB be the momentum after collision .Magnitude of vector OA and OB are equal and perpendicular to wall and equal and opposite .∆p = mvsin30° -(-mvsin30°) = 2mvsin30°F×t = 2mvsin30°F=2mvsin30°/tF =2×0.5×12×0.5/0.25F=24N

BalaJee

7 Years ago

12sin30+12sin30=12m/sec.

V=u+at

12=0+a×0.25

a=12/0.25

Hence,Force=mass × acceleration

12/0.23 ×0.5 N

25 N.

Rishi Sharma

5 Years ago

Dear Students,
Please find below the solution to your problem.

Favg=m(vf−vi)/tvf=0m/s
vi=12m/s
Therefore,
Favg=m(vf−vi)/t
=0.5(0−12)/0.25
=−24N

Thanks and Regards

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