A 6000-kg dump truck has a 1500-kg stone block sitting in its bed when the operator accidentally raises the bed to an angle of 30°. At this angle, the cables holding the block in place break so that the block slides down the bed and impacts the tailgate. Neglect the friction between the block and the bed and assume that the truck can roll freely. Determine the speed of the truck and the block immediately before the block hits the tailgate. Assume a plastic impact. The speed of the truck and the block immediately before the block hits the tailgate is m/s and m/s, respectively.

A) v_b = 4.43 m/s and v_t = 0 m/s

B) 0.886 m/s

Explanation:

A) We are given;

Mass of truck; M_t ; 6000 kg

Mass of block; M_b; 1500 kg

Angle at which operator incidentally raises the bed; θ = 30°

Now,using principle of conservation of energy, we have;

PE1 + KE1 = PE2 + KE2

Now, before the truck hits the toll gate, we have;

m_b(gh) = (1/2)m_b(v_b)²

Now, due to the angle, h = 2 sin 30 = 2 x 0.5 = 1.

Thus, plugging in the relevant values;

1500(9.8 x 1) = (1/2)•1500•(v_b)²

1500 will cancel out to give;

9.8 = (1/2)•(v_b)²

(v_b)² = 9.8 x 2

(v_b)² = 19.6

v_b = √19.6

v_b = 4.43 m/s

Meanwhile, v_t = 0 m/s

B) from conservation of linear momentum, the block and truck will have same velocity(v') after the impact. Thus;

m_b•v_b + m_t•v_t = (m_b + m_t)v'

Plugging in the relevant values to get ;

(1500 x 4.43) + (6000 x 0) = (1500 + 6000)v'

6645 = 7500v'

v' = 6645/7500 = 0.886 m/s

answer: 500 wph.

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