A boomerang (figure 4a) is modelled as pair of point masses, A and C, of mass m_

Question

A boomerang (figure 4a) is modelled as pair of point masses, A and C, of mass m_A = m_C = 0.1 kg separated by 0.2 m (figure 4b). At the instant shown in figure 4(b), the boomerang collides with a small pebble of mass m_P = 0.1 kg located at r_P|B = 0.05i - 0.15j m relative to B. Just before collision, the centre of mass of the boomerang at B is travelling at v^-_B = 10i + 5j ms^-1 relative to the inertial frame, and with angular velocity, omega^- = theta^- = 40 s^-1, and orientation, theta = pi/2, relative to the translating, but non-rotating, xy-frame. Meanwhile, the small pebble (of negligible dimensions) is travelling at v^-_P = 4i - 5j ms^-1 relative to the inertial frame. Just after the collision, the velocity of the pebble is observed to be v^+_P = 6i - 3j ms^-1 relative to the inertial frame. Since the force due to gravity is not impulsive, it is neglected in the analysis. (a) Draw the free-body diagram of only the pebble at the instant of the collision. Calculate the impulse describing the change in x and y linear momentum of the small pebble. Then draw the free-body diagram of only the boomerang at the instant of the collision (b) Use the linear impulse acting on the boomerang to determine the mass-centred velocity of the boomerang, v^+_B, just after the collision (c) Use the angular impulse acting on the boomerang to determine the angular velocity of the boomerang, omega^+, just after the collision.

Explanation / Answer

a )

the momentum is Px = m vx

using

Jx = m vpx - m vpx

= 0.1 X 2

= 0.2 kg.m/s i

sameway

Jy = m vpy - m vpy

= 0.1 X 2

= 0.2 kg.m/s j

b )

we have from above data

mB = mA + mB

= 0.2 kg

using conservation of energy

0.2 ( vB - 10 i + 5 j ) + 0.1 ( 6 i - 3 j - 4 i + 5 j )

vB = ( 9 i - 1.5 j ) m/s

c )

w = vB X rpB / | r |2

w = 9 i - 1.5 j ( 0.05 i - 0.15 j ) / ( 0.152 - 0.052 )1/2

w = 3.2 i + 1.6 j

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