Question

A One-Dimensional Inelastic Collision Block 1, of mass n-8.70 ke. moves along a fiictioaless air track with speed ty-21.0 m/s. It collides with block 2, of mass mg510 kg, which was initially at rest. The blocks stick topether after the collision Before collision 2 After collision: Part A Find the manitade p of the total initial momentun of the two-block aystem Express your answer numerically You did not open hicts for this part. ANSWER: kg m/s Part B Find , the magnitude of the final velocity of the two-block system. Express your answer numerically You did not open hints for this part ANSWER m/s Part C Kinitial in the two-block system's kinetic energy due to the collision? What is the change AK Kol Express your auswer namerically in joules. You did not open hints for this part. ANSWER

Answer 1

1) we have

p1= m1*v1 = 8.70kg*21.0m/s = 182.7 kg-m/s

2) from the conservation laws,we have

m1*v1 = (m1 + m2)*V so V = m1*v1/(m1 + m2) = 8.70*21.0/(59.7)m/s = 3.06 m/s

3) K_i = 1/2*8.7*(21)²= 1918.35 J

Kf = 1/2(m1 + m2)*(3.06)² = 279.50

So ΔK = 279.50- 1918.35= - 1638.85 J

Pb. (7.50) The x coordinate of center of mass is

m1 m22+ m3r3 mim2m3

The x-coordinate for mass 1.31 kg =0

The x-coordinate for mass 1.50 kg = 0.5 m

The x-coordinate for mass 1.10 kg = 0.75 m

Hence, we get

1.31(0)1.50(0.5) 1.10(0.75)0.39 m 1.311.501.10

Pb. (7.26)  

Let v_1f be the final speed of the moving puck and

v_2f be the final speed of the puck at rest.

We have
v_1f= v1i(2*m1)/(m1+m2)
v_1f= 5.74(0.9)/(1.35)
v_1f= 3.83 m/s east

v_2f = v_1i (m1-m2)/(m1+m2)
v_2f = 5.74*(-0.45)/(1.35)
v_2f = 1.91 m/s west