Question

A wooden cube having mass 10 kg is dropped from the top of a building . After 1 s , a bullet of mass 20 g fired at it from the ground hits the block with a velocity of 1000 m/s at an angle of 30o to the horizontal moving upwards and gets imbedded in the block . The velocity of the block/bullet system immediately after the collision is: 17 m/s 27 m/s 52 m/s 10 m/s

Answer 1

Answer:

40ms

−1

upward

Explanation:

Time after which collision takes place

t=

u

h

=

100

100

=1sec

Initial velocity of the wooden block

u

1

=gt=10×1=10m/s

Initial velocity of the bullet

u

2

=u−gt=100−10×1=90m/s

m

1

u

1

−m

2

u

2

=(m

1

+m

2

)V

∴V=−40m/s

Hence the velocity immediately after collision will equal to 40 m/s in the upward direction .

Answer 2

Answer:

The velocity of the block/bullet system immediately after the collision is: 10m/s

Explanation:

• Step-1: Mass of wooden cube= m1 = 10kg

                    At top of building it was at rest, so initial velocity = u = 0

                    Acceleration = g = 9.8 m/s² , time = t = 1s

       From 1st equation of motion, final velocity = u1 = u + gt = 0 + (9.8 x 1)

                                                                                u1 = 9.8 m/s

• Step-2: Mass of bullet = m2 = 20g = 0.02 kg
•             Its velocity is 1000 m/s and at an angle of 30° with horizontal
• Vertical component of velocity = u2 = 1000(sin30°) = 1000×$\frac{1}{2}$
•                                                      
• u2 = 500 m/s
• Step-3: Before collision momentum of wooden cube = m1 × u1

                                                                              = 10 × 9.8 = 98 kg-m/s

   Before collision momentum of bullet= m2 × u2=0.02×500 =10 kg-m/s

   Before collision total momentum = m1u1 + m2u2 = 98 + 10 =108 kg-m/s

• Step-4: Suppose after collision velocity of the system = v      

                      After collision total momentum = (m1 + m2)v = (10 + 0.02)v

                                                                                                = 10.02v        

• Step-5: From momentum conservation law,

                                              m1u1 + m2u2 = (m1 + m2)v

                         Putting the values,    108 = 10.02 × v

                                                              v = 108/10.02 = 10.7 m/s

• Conclusion: The final velocity of the system = v = 10.7 m/s

                              The nearest option of the value is 10 m/s.

CODE:   #SPJ3