The figure shows an arrangement with an air track, in which a cart is connected

Question

The figure shows an arrangement with an air track, in which a cart is connected by a cord to a hanging block. The cart has mass m1 = 0.740 kg, and its center is initially at xy coordinates ( - 0.350 m, 0 m); the block has mass m2 = 0.350 kg, and its center is initially at xy coordinates (0, - 0.150 m). The mass of the cord and pulley are negligible. The cart is released from rest, and both cart and block move until the cart hits the pulley. The friction between the cart and the air track and between the pulley and its axle is negligible. (a) In unit-vector notation, what is the acceleration of the center of mass of the cart - block system? (b) What is the velocity of the corn as a function of time t, in unit - vector notation?

Explanation / Answer

we take our positive directions for m1 and m2 to the right and downward respectively

newtons second law yeilds that

m1 a = T

m2 a = m2 g - T

a = m2 g / (m1 + m2)

= (0.350) (9.8) / (0.740 + 0.350)

= (3.43) / (1.09)

= 3.1467 m / s2

in coordinate system given in the figure the accelerations of m1 and m2 are

a1 = + 3.1467 i

a2 = - 3.1467 j

the acceleration of centre of mass is

acom = (m1 a1) + (m2 a2) / (m1 + m2)

= 2.136(m/s2) i - 1.008(m / s2) j

The velocity of the center of mass is

vcom = vcom,0 + acomdt

= 2.136 t i – 1.008 t j

The center of mass of the system is initially located at

xcom,0 = (m1 x1,0 + m2 x2,0 ) / (m1 + m2)

The center of mass of the system is

xcom = xcom,0 + vcom dt

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