Mass Mi. with initial speed vo, slides to the east on ice and collides with mass

Question

Mass Mi. with initial speed vo, slides to the east on ice and collides with mass M. imitially at rent Ate the collision Mi moves to the northeast with velocity v at an angle 0 he south Let A1,-4 kg, M2 = 3 kg, vo-5 m/s, and -30. with the east direction and M A) Find v : moves to a) 6.22 m/s. b) 5.87 m/s,) 6.71 ms) s77 m/s B) Find 12 4.63 m/s. b) 4.90 m/s, c) 385 m/s. d 5.75 m/s C) Find the initial kinetic energy in the system. a) 63.5 J a) 134J a) 4040J b) 50J c 28 d 38.5 b) 88.9 J, )116J )988 b) 4780 J. c) 8340 d the final kinetic energy in the system. (Note: the collision is not elastic) find the average force on M2 during the E) Ifthe collision lasts for a time t=0.005 s, d) 2310

Explanation / Answer

momentum before collision = momentum after collision

M1*voi = M1*(v1*cos30i + V1*sin30 j) - M2*v2 j

4*5 i = 4*(v1*cos30 i + V1*sin30 j) - 3*V2 j

20i = 4*v1*cos30 i + ( 4v1*sin30 - 3*v2) j

4*v1*cos30 = 20

v1 = 5.77 m/s

( 4*v1*sin30 - 3*v2) = 0

4*v1*sin30 = 3*v2

4*5.77*sin30 = 3*v2

v2 = 3.85 m/s

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A)

v1 = 5.77 m/s <<------------ANSWER

OPTION (d)

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B)

B) v2 = 3.85 m/s

OPTION ( C )

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C)

initial kinetic energy Ki = (1/2)*m*vo^2 = (1/2)*4*5^2 = 50 J

OPTION ( b )

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D)

final kinetic energy Kf = (1/2)*M1*v1^2 + (1/2)*M2*v2^2

Kf = (1/2)*4*5.77^2 + (1/2)*3*3.85^2

Kf = 88.9 J

OPTION ( b)

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E)

force on M2 , F = M2*V2/t = 3*3.85/0.005 = 2310 N <<-ANSWER

OPTION ( d )

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