Block B (mass 7.50 kg ) is at rest at the edge of a smooth platform, 2.60 m above the floor. Block A (mass 4.00 kg ) is sliding with a speed

Question

Block B (mass 7.50 kg ) is at rest at the edge of a smooth platform, 2.60 m above the floor. Block A (mass 4.00 kg ) is sliding with a speed of 8.00 m/s along the platform toward block B. A strikes B and rebounds with a speed of 2.00 m/s. The collision projects B horizontally off the platform.

What is the speed of B just before it strikes the floor?

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Thiên Di 4 years 2021-09-01T01:07:48+00:00 1 Answers 119 views 0

Answers ( )

  1. Neala
    0
    2021-09-01T01:09:30+00:00
    Reply

    Answer:

    speed of block B after collision is 5.33 m/s

    speed of block B before strike speed 8.9 m/s

    Explanation:

    given data

    mass = 7.50 kg

    height = 2.60 m

    speed =8.00 m/s

    rebound speed = 2.00 m/s.

    solution

    we will apply here conservation of momentum that is

    m1v1  + m2v2  = m1vr + m2v    ……………..1

    4 × 8 + 7.50 × 0 = 4 × (-2) + 7.50 × v  

    solve it we get

    v = 5.33 m/s

    so here speed of block B after collision is 5.33 m/s

    and

    and now by conservation of energy

    we get here strike speed that is

    PE(i) + KE(i) = PE(f) + KE(f)   …………..2

    m×g×h + 0.5 × m × (v(i))² = 0 + 0.5 ×m×(v(f))²  

    m ( g×h + 0.5×(v(i))²) = 0.5 ×m×(v(f))²  

    9.8 × 2.60 + 0.5 × 5.33² = 0.5  × (v(f))²  

    solve it we get

    v(f)² = \frac{39.7}{0.5}  

    v(f) = \sqrt{\frac{39.7}{0.5}}

    v(f) = 8.9 m/s

    so speed of block B before strike speed 8.9 m/s

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