A 26.5-g object moving to the right at 20.5 cm/s overtakes and collides elastically with a 12.5-g object moving in the same direction at 15.

Question

A 26.5-g object moving to the right at 20.5 cm/s overtakes and collides elastically with a 12.5-g object moving in the same direction at 15.0 cm/s. Find the velocity of each object after the collision. (Take the positive direction to be to the right. Indicate the direction with the sign of your answer.)

in progress 0
Bình An 3 years 2021-07-27T14:19:24+00:00 1 Answers 21 views 0

Answers ( )

  1. ngocdiep
    0
    2021-07-27T14:20:53+00:00
    Reply

    Answer:

    v₁ =0.19 m/s and v₂ = 0.18 m/s

    Explanation:

    By conservation of energy and conservation of momentum we can find the velocity of each object after the collision:

    Momentum:

    Before (b) = After (a)

    m_{1}v_{1b} + m_{2}v_{2b} = m_{1}v_{1a} + m_{2}v_{2a}

    26.5 kg*0.205 m/s + 12.5 kg*0.150 m/s = 26.5 kg*v_{1a} + 12.5 kg*v_{2a} 7.31 kg*m/s = 26.5 kg*v_{1a} + 12.5 kg*v_{2a}     (1)      

    Energy:

    Before (b) = After (a)                        

    \frac{1}{2}m_{1}v_{1b}^{2} + \frac{1}{2}m_{2}v_{2b}^{2} = \frac{1}{2}m_{1}v_{1a}^{2} + \frac{1}{2}m_{2}v_{2a}^{2}          

    26.5 kg*(0.205 m/s)^{2} + 12.5 kg*(0.150 m/s)^{2} = 26.5 kg*v_{1a}^{2} + 12.5 kg*v_{2a}^{2}              

    1.40 J = 26.5 kg*v_{1a}^{2} + 12.5 kg*v_{2a}^{2}   (2)    

    From equation (1) we have:

    v_{2a} = \frac{7.31 kg*m/s - 26.5 kg*v_{1a}}{12.5 kg}   (3)

    Now, by entering equation (3) into (2) we have:  

    1.40 J = 26.5 kg*v_{1a}^{2} + 12.5 kg*(\frac{7.31 kg*m/s - 26.5 kg*v_{1a}}{12.5 kg})^{2}    (4)

    By solving equation (4) for v_{1a}, we will have two values for

    v_{1a} = 0.16                        

    v_{1a} = 0.21  

    We will take the average of both values:

    v_{1a} = 0.19 m/s

    Now, by introducing this value into equation (3) we can find v_{2a}:

    v_{2a} = \frac{7.31 kg*m/s - 26.5 kg*0.19 m/s}{12.5 kg}

    v_{2a} = 0.18 m/s

    Therefore, the velocity of object 1 and object 2 after the collision is 0.19 m/s and 0.18 m/s, respectively.

    I hope it helps you!    

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )