A 0.400 kg block sliding on a frictionless floor with a velocity of 3 00 m/s Eas

Question

A 0.400 kg block sliding on a frictionless floor with a velocity of 3 00 m/s East has an elastic collision with a 0.600 kg block that has a velocity of 5.5O m/s West. the 0.400 kg block then hits a spring with a spring constant of 332 N/m, slowing to a stop, while the 0.600 kg block has a completely inelastic collision with a 0.800 kg block that has a velocity of 2.20 m/s West. What are the magnitudes and directions of the velocities of the 0.400 kg block and the 0.600 kg block right after the elastic collision? By how many meters is the spring compressed? What is the magnitude and direction of the velocity of the 0.600 kg block and 0.800 kg block after the completely inelastic collision?

Explanation / Answer

a). In this case we have to use conservation of Linear momentum at 1st.

let m1=0.4kg , m2=0.6g , u1=3m/s(east), u2=5.5m/s(west)
velocity after collision of m1 be v1 & of m2 be v2

Using the momentum conservation theorem we get

m1u1-m2u2=m1v1+m2v2 (here east is taken as '+ve' direction & west as'-ve' direction)   eqn-1

Again we know

     Coefficent of Restitution(e)=Relative velocity after collision/Relative velocity before collision

according to question this colision is elastic, so e=1.

u1+u2=relative velocity before collision( as the block were moving in opposite direction)

v2-v1=relative velocity after collision (we have taken that after collision they move in same direction and v1<v2)

so we have

u1+u2=v2-v1 eqn-2

Multiplying eqn-2 by m1 & adding with eqn-1 we get

2m1u1+(m1-m2)u2=(m1+m2)v2

or, v2=(2m1u1+(m1-m2)u2)/(m1+m2)

or, v2=((2*0.4*3)-(0.2*5.5))/1 (m1=0.4kg , m2=0.6g , u1=3m/s, u2=5.5m/s)

or, v2=1.3 m/s (as sign '+ve' it is in east direction)

Multiplying eqn-2 by m2 & subtracting from eqn-1 we get

(m1-m2)u1-2m2u2=(m1+m2)v1

or, v1=((m1-m2)u1-2m2u2)/(m1+m2)

or, v1= ((-0.2)*3 - 2*0.6*5.5)/1    (m1=0.4kg , m2=0.6g , u1=3m/s, u2=5.5m/s)

or, v1= -7.2 m/s('-ve' indicates it is in west direction)

b)

let elastic constat of spring be k=332 N/m

Now block of mass m1 lose all of its kinetic energy to contract the spring and it store as the potential energy of spring.

Let the maximum contraction of spring be 'x'

Then potential energy stores in spring be '(1/2)*kx2'

The kinetic energy of mass m1 be '(1/2)*m1v12'

So we have

(1/2)*kx2=(1/2)*m1v12

or, x=sqrt(m1v12/k)

or, x=sqrt(0.4*7.22/332) 9 (m1=0.4kg, v1= 7.2, k=332 N/m)

or, x=0.25m= 25 cm

c) For 3rd block m3=0.8kg, velocity v3=2.2m/s(west direction)

& For 2rd block m2=0.6kg, velocity v2=1.3m/s(east direction)

In this case collision is completely inelastic so they will combine and move with common velocity after collision.

Let the common velocity be 'V'

Same as before in this case we have

(m2+m3)V=m2v2-m3v3 (sign convention is as before)

or, V=(m2v2-m3v3)/(m2+m3)

or, V=(0.6*1.3 - 0.8*2.2)/1.4     (m3=0.8kg, v3=2.2m/s, m2=0.6kg, v2=1.3m/s)

or, V= -0.7 m/s('-ve' indicates it is in west direction)

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