A track consists of a frictionless arc XY, which is a quarter-circle of radius R, and a rough horizontal section YZ. Block A of mass M is released from rest at point X, slides down the curved section of the track, and collides instantaneously and inelastically with identical block B at point Y. The two blocks move together to the right, sliding past point P, which is a distance l from point Y. The coefficient of kinetic friction between the blocks and the horizontal part of the track is ì. Express your answers in terms of M, R, ì, R, and g.

The question is not complete, so i have attached an image with the complete question.

A) speed of block a before it hits block B;v = √2gR

B) speed of combined blocks after collision;V_ab = ½√2gR

C) kinetic energy lost = ½MgR

D) temperature change; ∆t = ½gR/c

E) Additional thermal energy;W_f = 2μMgL

Explanation:

A) From conservation of energy, potential energy = kinetic energy.

Thus; P.E = K.E

So, mgr = ½mv²

m will cancel out to give;

gr = v²

So, v = √2gR

B) from conservation od momentum, momentum before collision = momentum after collision.

Thus;

M_a•V_a = M_ab•V_ab

Where;

M_a is the mass of block A

V_a is speed of block A

M_ab is the mass after collision

V_ab is speed after collision

So, making V_ab the subject, we have;

V_ab = M_a•V_a/M_ab

Now from answer in a above,

V_a = V = √2gR

Also, M_a = M and M_ab = 2M because it's the sum of 2 masses after collision.

Thus;

V_ab = (M√2gR)/(2M)

M will cancel out to give;

V_ab = ½√2gR

C) From the work-kinetic energy theorem, the net work done on the object is equal to the change in the kinetic energy of the object.

Thus;

W_net = K_f - K_i = ½m_ab•v_ab²- ½m_a•v_a²) = ∆K.

We have seen that:

M_a = M and M_ab = 2M

V_a = √2gR and V_ab = ½√2gR

Thus;

∆K = ½(2M•(½√2gR)²) - ½(M•(√2gR)²)

∆K = ½MgR - MgE

∆K = -½MgR

Negative sign means loss of energy.

Thus kinetic energy lost = ½MgR

D) The formula for heat energy is given by;

Q = m•c•∆t

Thus, change in temperature is;

∆t = Q/Mc

Q is the heat energy, thus Q = ½MgR

Thus;∆t = ½MgR/(Mc)

M will cancel out to give;

∆t = ½gR/c

E) Additional thermal energy is gotten from;

Work done by friction;W_f = F_f x d

Where;

F_f is frictional force given by μF_n. F_n is normal force

d is distance moved

Thus;

W_f = μF_n*d

Mass after collision was 2M,thus, F_n = 2Mg

We are told to express distance in terms of L

Thus;

W_f = μ2Mg*L

W_f = 2μMgL

hey , the answer is c

3.2x10^7