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Answer:
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
