A 3.00-kg steel ball strikes a massive wall at 10.0 m/s at an angle of ϴ = 60.0° with the plane of t

A 3.00-kg steel ball strikes a massive wall at 10.0 m/s at an angle of ϴ = 60.0° with the plane of the wall. It bounces off the wall with the same speed and angle. If the ball is in contact with the wall for 0.200 s, what is the average force exerted by the wall on the ball?

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[@habtamumullubiadgo2807]

This problem has been solved!
See the answer
A 0.500 kg sphere moving with a velocity (2.00i - 2.90j + 1.00k) m/s strikes another sphere of mass 1.50 kg moving with a velocity (-1.00i + 2.00j - 2.50k) m/s.
(a) If the velocity of the 0.500 kg sphere after the collision is (-0.90i + 3j - 8.00k) m/s, find the velocity of the 1.50 kg sphere and identify the kind of collision (elastic, inelastic, or perfectly inelastic).
(b) If the velocity of the 0.500 kg sphere after the collision is (-0.250i + 0.775j - 1.63k) m/s, find the final velocity of the 1.50 kg sphere and identify the kind of collision.
(c) What if? If the velocity of the 0.500 kg sphere after the collision is (-1.00i + 3.10j + ak) m/s, find the value of a and the velocity of the 1.50 kg sphere after an elastic collision. (Two values of a are possible, a negative value and a positive value. Report each with their corresponding final velocities.)

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[@teiermyler4926]

I just did 30*cos30 to find the x component of momentum. Then multiplied that momentum by 2 because the impulse of the wall has to both stop the incoming x momentum, and then supply that same momentum again outward. And dividing that value by 0.2 s gets a force of 259.808N

[@olbibulbi4348]

where did you get 30 from???

[@noyamanu1242]

@@olbibulbi4348 Right angle marked by the black line makes a 90 degree angle, and 90-60 is 30.

[@noyamanu1242]

@@olbibulbi4348 Unless you mean the actual 30 and not cos30, in which that was just the speed (10 m/s) times the mass (3kg)

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What is the diameter of a 3 kg steel ball???

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how you resolved that components. actually x component is cos:O

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