The accompanying drawing shows a collision between two pucks. Puck A has a mass

Question

The accompanying drawing shows a collision between two pucks. Puck A has a mass of 0.02 kg (m_A = 0.02 kg) and is moving along the x-axis with a velocity of +6.0 m/s. It makes a collision with another puck B, initially at rest and has a mass of 0.05 kg (m_B = 0.05 kg). The collision is not head-on. After the collision the two pucks fly apart along the directions shown in figure. (a) What is the total initial momentum of the m_A + m_B system? Please clearly provide magnitude and direction (b) What is the total x-component of the final momentum of the m_A + m_B system? Please clearly provide magnitude and direction. (c) What is the total y-component of the final momentum of the m_A + m_B system? Please clearly provide magnitude and direction. (d) Show that v^f_A = Squareroot 3v^f_B, where v^f_A and v^f_B are the magnitude of the final velocities of the puck A and puck B respectively. (e) Find the velocities of puck A and puck B after the collision.

Explanation / Answer

Ma=0.02 kg

Mb.05kg

UA(initial velocity of ball a)=6m/s

UB(initial velocity of ball b)=0m/s

A)initial momentum of puck a and b is=(0.02*6)+(0.05*0)=0.12 kgm/s along positive x axis

B)final momentum of puck a is =mav1cos60°i+(mav1sin60°)j

Final momentum of puck b is =mbv2 cos30°i+(mvbsin30°*(-j))

Total final momentum of puck a and puck b is.

=(0.01v1+0.043v2)I+(0.017v1-(0.025v2)j

By law of conservation of momentum initial momentum is equal to final momentum

0.12i+0j=(0.01v1+0.043v2)I+(0.017v1-0.025v2)j

C)X component of total final momentum is 0.12 kg m/s

D)Y component of total final momentum is 0 kgm/s

E) from conservation of momentum

0.02sin60v1-0.05v2sin30=0

Since there is no momentum along y axis initiTially

V1=(2.5/3)v2

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