lllll llL a triangle with the angies and Compute the mass of the uniform b 2. El

Question

lllll llL a triangle with the angies and Compute the mass of the uniform b 2. Elastic collision of two particles: The particie has a mass of 0.1kg and has a first velocity of 1 e The second particle has the same mass and has the same velocity (in the opposite direction a) What is the velocity of the two particles after the collision assume that the incoming particle moves to the right)? b) Compute the velocity of the center of mass, before and after the collision. c) Assume that the duration of the collision is 0.01 seconds. What will be the impuls on each particle? d) What will the kinetic energy of each particle before and after the collision? 3. Consider the following torsional pendulum. A wire of lenth 9.8m is attached to a mass

Explanation / Answer

(a) let v1= 1m/s

v2= -1 m/s

m1= m2= 1kg

vf1= (m1-m2)v1/(m1+m2)+ 2 m2 v2/(m1+m2)

vf1= (0.1-0.1)*1/(0.1+0.1) + 2 *0.1*(-1)(0.1+0.1)

vf1=1 m/s.

vf2= (m2-m1)v2/(m1+m2)+ 2 m1 v1/(m1+m2)

vf2= (0.1-0.1)*1/(0.1+0.1) + 2 *0.1*(1)(0.1+0.1)

vf2=1 m/s.

(b)before collision, Vcm= (m1v1+m2v2)/(m1+m2)

Vcm=(0.1*1+ (-0.1)*1)/(0.1+0.1)

Vcm=0

after collision,

Vcm= (m1v1f+m2v2f)/(m1+m2)

Vcm=(-0.1*1+ 0.1*1)/(0.1+0.1)

Vcm=0.

(c) for first particle, impulse= change in momentum= m1 (v1f-v1)

i1=0.1 (-1-1)

i1=-0.2 kgm/s

for second particle, impulse= change in momentum= m2 (v2f-v2)

i2=0.1 [1-(-0.1)]

i1=0.2 kgm/s

(d)beforec collison, KE1= 1/2 m1v12= 1/2 (0.1)(12)=0.05j

KE2= 1/2 m2v22= 1/2 (0.1)(-1)2=0.05j

after collision;

KE1f= 1/2 m1v1f2= 1/2 (0.1)(-1)2=0.05j

KE12= 1/2 m2v2f2= 1/2 (0.1)(12)=0.05j

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