An air track system has two carts A and B of mass mA = 0.3 kg and mB = 0.6 kg on

Question

An air track system has two carts A and B of mass mA = 0.3 kg and mB = 0.6 kg
on a frictionless track. Cart A is moving at a speed of 0.5 m/s towards cart B,
which is stationary. An elastic collision occurs. Answer the following questions.
a) Find the speed and direction of cart A after the collision.
b) Find the speed and direction of cart B after the above collision.
c) What is the kinetic energy of cart B after the collision?
d) What is the speed of the center of mass of the system?
e) Assuming cart A is in contact with cart B for 2.0 milliseconds, find the average
force acting on cart A during the collision.

Explanation / Answer

Mass of the cart A = Ma = 03 kg Mass of the cart B = M b = 0.6 kg intial speed of the cart A = Ua = 0.5 m/s intial speed of the cart b = Ub = 0 m/s (rest ) a) Apply consrevation of momentum to this system    we have ,            Ma Ua +Mb Ub = Ma Va + Mb Vb           for elastic collision ,                co-effeicent of restitution is = 1                       e = Vb - Va / Ua - Ub= 1                    Vb - Va =Ua - Ub                                  = 0.5 m/s -0                           Vb = 0.5 + Va            thus,                 Ma Ua +Mb Ub = Ma Va + Mb Vb                  (0.3)(0.5)+ (0.6)(0) = (0.3)(Va)+ (0.6)(0.5 + Va)                              0.15 = 0.3 Va+ 0.3 + Va (0.6)                                      = 0.9 Va + 0.3                               Va = -0.167 m/s         so, Speed of the cart A after the collision takes place is :                           Va = 0.167 m/s                     direction : opposite to its intial direction         (b)                  thus, Vb = 0.5 + Va = 0.5 - 0.167 m/s                                  = 0.33 m/s                            Dorection is same as intial direction of it     (c)                  Kinetic energy of the cart B after the collision                       = 1/2 Mb Vb2                        = 0.5 (0.6)(0.33)2                       = 0.0332667 J    (d)              For speed of the center of the Mass system                 Vcm = Ma Va + Mb Vb /( Ma + Mb )                           = (0.3)(-0.167)+ (0.6)(0.33) / (0.3+0.6)                           =0.1643 m/s (e)          Average force acting on the cart A :                 F = M a (Ua - Va )                     = (0.3)(0.5- (-0.167))                     = 0.2001 N (e)          Average force acting on the cart A :                 F = M a (Ua - Va )                     = (0.3)(0.5- (-0.167))                     = 0.2001 N                                              

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