Question

• If a block of mass m at speed v collides inelastically with a block of mass M at rest, what is the final velocity of each block? What fraction of the initial energy is lost?

• Two identical circular hockey pucks of radius r traveling towards each other at the same speed v collide elastically. Their centers are separated by a vertical distance d, which is less than r. After the collision, with what angle relative to the initial path of the pucks will each one travel along?

• Explain the physics of Newton’s Cradle.

Answer 1

Linear momentum=mv,where m is mass of the object and v is its velocity

Since there are no external forces horizontally,linear momentum is conserved in horizontal direction.

Let right direction be taken as positive and let initial motion of m be in rightward direction.

Initially, mass m is moving with velocity v towards right, mass M is at rest. So,initial linear momentum=m*(+v)+M(0)=mv

Finally, since there is inelastic collision, both move with same velocity.Let they be moving with velocity v₁ towards right. So,Final momentum=(m+M)(+v₁)=(m+M)v₁

Using momentum conservation,initial momentum=final momentum=>mv=(m+M)v₁=>v₁=mv/(m+M)

So,final velocity of each block=mv/(m+M)

Kinetic energy=1/2 (mass)(velocity)²

Initially,mass m is moving with speed v while mass M is at rest,So, Kinetic energy=1/2mv²+1/2M(0)²=1/2mv²

Finally,both masses are moving with velocity=mv/(m+M),So, Kinetic energy=1/2 m [mv/(m+M) ]²+1/2M[ mv/(m+M) ]^{2 =} 1/2(m+M)[ mv/(m+M) ]² =(mv)²/[2(m+M)]

Loss of kinetic energy=initial kinetic energy-final kinetic energy=1/2mv² - (mv)²/[2(m+M)]=1/2mv²[1-m/(m+M)]

=1/2mv²(M) / (m+M)

So,fraction of initial energy lost=loss in kinetic energy / initial kinetic energy

=>fraction of initial kinetic energy lost=0.5mv²(M) / [(m+M)*0.5mv²]= M / (m+M)