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anastassius [24]
3 years ago
5

A 10kg sphere hits a stationary 8kg sphere. After the collision the 8kg sphere moves off in the positive direction at 4m/s. If t

he 10kg sphere was originally moving at 10m/s, what is its velocity after the collision?
Physics
1 answer:
Bumek [7]3 years ago
7 0

M = mass of the first sphere = 10 kg

m = mass of the second sphere = 8 kg

V = initial velocity of the first sphere before collision = 10 m/s

v = initial velocity of the second sphere before collision = 0 m/s

V' = final velocity of the first sphere after collision = ?

v' = final velocity of the second sphere after collision = 4 m/s

using conservation of momentum

M V + m v = M V' + m v'

(10) (10) + (8) (0) = (10) V' + (8) (4)

100 = (10) V' + 32

(10) V' = 68

V' = 6.8 m/s

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maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

       The length of the track is l =  1.0 \ km  =  1000 \  m

       The speed of  A is  v__{A}} =  20 \ m/s

       The uniform acceleration of  B is  a__{B}} =  0.333 \ m/s^2

  Generally the time taken by go-cart  A is mathematically represented as

              t__{A}} = \frac{l}{v__{A}}}

=>          t__{A}} = \frac{1000}{20}

=>           t__{A}} =  50 \  s

  Generally from kinematic equation we can evaluate the time taken by go-cart B as

             l =  ut__{B}} + \frac{1}{2}  a__{B}} * t__{B}}^2

given that go-cart B starts from rest  u =  0 m/s

So

            1000 =  0 *t__{B}} + \frac{1}{2}  * 0.333  * t__{B}}^2

=>         1000 =  0 *t__{B}} + \frac{1}{2}  0.333  * t__{B}}^2            

=>         t__{B}} =  77.5 \  seconds  

 

Comparing  t__{A}} \  and  \ t__{B}}  we see that t__{A}} is smaller so go-cart A is  faster

   

       

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Answer:

C. 30.6m

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