) Mass m1 = 2kg is initially moving with speed v1=0.90 m/s on a frictionless sur

Question

) Mass m1 = 2kg is initially moving with speed v1=0.90 m/s on a frictionless surface . Mass m2 = 1 kg is intially at rest. Mass m1 collides and sticks to mass m2 . They move together with velovity vf toward a spring of elastic constant K= 900 N/m

a- What is the speed of the two masses after collision ?
b-What is the magnitude of the change in momentum of mass m1 due to the collision with mass m2 ?
c-what is the change in kinetic energy of the two masses due to the collision ?
d-What is the maximum compression of the spring ?

Explanation / Answer

A. using momentum conservation

m1v1 + m2v2 = (m1+m2)*V

V = (2kg*0.90 m/sec + 1kg*0)/(2 + 1) = 0.6 m/sec

Speed of the two masses = 0.6 m/sec

B. P1i = m1*v1 = 2*0.9 = 1.8 kg-m/sec

P1f = m1*V = 2*0.6 = 1.2 kg-m/sec

dP1 = 1.2 - 1.8 = -0.6 kg-m/sec

C.

KEi = 0.5*m1*v1^2 + 0.5*m2*v2^2

KEi = 0.5*2*0.9^2 + 0.5*1*0^2

KEi = 0.81 J

KEf = 0.5*(m1+m2)*V^2

KEf = 0.5*3*0.6^2 = 0.54 J

dKE = 0.54 - 0.81 = -0.27 J

D.

KE of block = PE of spring at maximum compression

0.5*(m1+m2)*V^2 = 0.5*k*x^2

x = sqrt(2*KEf/k)

k = 900 N/m

x = sqrt(2*0.54/900) = 0.0346 m

maximum compression = 3.46 cm

let me know if you have any doubt.

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