PHYSICS 81LOCUS

An elevator platform is going up at a speed 20 m/s and during its upward motion a small ball of mass 50 gm fallingin downward direction at a speed of 5 m/s. Find the speed with which the ball rebounds after an elastic collision withthe platform.

As the mass of the platform is very large compared to that of the ball, we can assume that the speed ofthe platform remains unchanged. If v be the velocity of the ball just after the impact, as show in figure 6.63.

m s m s

v

m s

Using equation (6.40), we have

velocity of separation = e.velocity of approachvelocity of separation = velocity of approach [ 1]e

20 20 5v

45v m/s

Aneutron moving at a speed u undergoes a head-on elastic collision with a nucleus of mass number A at rest. Find theratio of kinetic energies of neutron after and before collision.

If mbe the mass of neutron then mass of the nuclei will be mA. Let 1v be the speed of the neutron and 2vbe that of the nuclei after the collision as shown in figure 6.64.

u

m mA m mA

v v

Applying conservation of linear momentum, we have

in finP P

1 2mu mv mAv

1 2u v Av ...(i)

Using equation (6.40), we have

.sep approachv v

PHYSICS 82LOCUS

Solving equations (i) and (ii) , we get

111

Av uA

Thus KE of neutron after collision is

21

12fk mv

221 1

2 1AmuA

The KE of electron before collision was

212ik mu

Therefore, required ratio is

211

f

i

k Ak A

When colliding bodies exert forces on each other which are not along the line parallel to the direction of their initialmotion, directions of motion of bodies change due to the collision. Such type of collisions are called obliquecollisions. Consider the situations shown in figures 6.65. In figure 6.65(a) two smooth spherical balls A and B,initially

v

v

u

u

NN –

A

A

A

B

B

B

m

m

f jP =P +J

P iJ = N dt•••

A

f jP =PP i

J = – N dt••

•

moving with speeds u and u collide obliquely and then move with speed v and v respectively, along directions

PHYSICS 83LOCUS

contact force on each sphere act perpendicularly away from the surface of contact, as shown in figure 6. , and dueto the spherical geometry of each body, normal contact force on each body passes through its centre. Normalcontact force on each body imparts impulse to the body along the direction of its action. If impulse on a body isalong the line parallel to the direction of initial momentum of the body then final momentum is also along the same linebut if the impulse is not along that line, as in this case, the final momentum of the body is along a different direction.In figures 6. 65(b) and 6.65 (c) final momenta of the bodies A and B, respectively, are obtained by adding impulseon each body to the initial momentum of the body.

N NC C

Common tangentialsurface, called assurface of contact

Common tangential

surface, called as

surface of contact

A similar case is illustrated in figure 6.67. The ball A strikes the horizontal surface obliquely and receives

N

A

A B

N–

u

v

A

J = N dt••

•

Pf

Pi

impulse along the direction perpendicularly away from the surface which changes the direction of the motion of theball as shown in figures 6.67(a) and 6.67(b).

Now, let us discuss the oblique collision in a different way which makes the solution of problems involvingsuch collisions very simple. The situation shown in figure 6.65 is analyzed again, as shown in figure 6.68 Situationsof balls A and B just before the collision (figures 6.68(a), (b), (c)), during the collision (figure 6.68(d)) and just afterthe collision