Question

A mass is moving at 8 m/s in the +x direction and it collides in a perfectly elastic collision with a mass of 4 kg moving in the -x direction. The collision takes places in 0.20 seconds and after the collision the mass that was moving in the +x direction is moving in the -x direction at 9 m/s and the mass that was moving in the -x direction is moving in the +x direction at 11 m/s. What is the magnitude of the average force, in Newtons, on the first mass which was originally moving in the +x direction before the collision?

Answer #1

**given**

**u1 = 8 m/s**

**m2 = 4 kg**

**after the collision,**

**v1 = -9 m/s**

**v2 = 11 m/s**

**let m1 is the mass of the first body and u2 is the
initial velocity of the second body.**

**we know,**

**v1 = ( (m1 - m2)*u1 + 2*m2*u2 )/(m1 + m2)**

**(-9) = ( (m1 - 4)*8 + 2*4*u2 )/(m1 + 4)
----(1)**

**we know,**

**v2 = ( (m2 - m1)*u2 + 2*m1*u1 )/(m1 + m2)**

**11 = ( (4 - m1)*u2 + 2*m1*8 )/(m1 + 4)
----(2)**

**on solving above two equationms we get**

**m1 = 5.4 kg**

**u2 = -12 m/s**

**magnitude of average force acting on first body, |F_avg|
= m1|a|**

**= m1*(u1 - v1)/t**

**= 5.4*(8 - (-9))/0.2**

**= 459 N
<<<<<<<<<<<---------------Answer**

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