ADDITIONAL PROBLEMS Question 6.2a: Two blocks slide on a collision course across

Question

ADDITIONAL PROBLEMS Question 6.2a: Two blocks slide on a collision course across a frictionless surface, as in the figure. The resulting collision is inelastic. The first block has mass M 1.00kg and is initially sliding due north at aspeed of V-895 m/s The second block has mass m#7.00 x 10-2 kg and is initially sliding at a speed of 1.5 m/s directed at an angle ?= 295·south of east. Immediately after the inelastic colision, the second block is observed sliding at ? . 4.95 m/s in a direction of ? . 215, north of east. Block 2 M Block 1 Determine the components of the first block's velocity after the collision. m/s Question 6.2b: During a fireworks display,a 0.290-kg bottle rocket approaches the top of ts trajectory and explodes into two pieces as in the figure The first piece (Piece 1) has mass 0.190 kg and the second piece (Piece 2) has mass 0.100 kg. Immediately after the explosion, the first piece is observed traveling to the left at a speed of 4.10 m/s and the second piece is observed traveling at an angle ? 37.0° above the horizon at a speed of 170m/s Piece 1 Bottle rocket Determine the components of the bottle rocket's velocity immediately before the explosion m/s m/s Need Help?Resd It

Explanation / Answer

6.2 a

according to the figure, there is no friction anywhere and the collision is inelastic

M = 1 kg

U = 8.95 m/s j

m = 7*10^-2 kg

u = 11.5(cos(29.5)i - sin(29.5)j)

also,

V = Vf(cos(a)i + sin(a)j)

v = vf(cos(21.5)i + sin(21.5)j)

hence from conservation of momentum

MU + mu = MV + mv

1*8.95 - 7*10^-2*sin(29.5) = 1*Vf*sin(a) + 7*10^-2*4.95*sin(21.5)

7*10^-2*11.5*cos(29.5) = Vf(cos(a)) + 7*10^-2*4.95*cos(21.5)

=>

[8.95 - 0.07sin(29.5) - 0.07*4.95*sin(21.5)]^2 + [0.07*11.5*cos(29.5) - 0.07*4.85*cos(21.5)]^2 = Vf^2

Vf = 8.796 m/s

a = 87.6397240 deg

Vxf = Vf*cos(a) = 0.362245113459545 m/s

Vyf = Vf*sin(a) = 8.788537675614073 m/s

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