Question

A bullet of mass 20g is fired with a speed of 100m/s into a wooden block of mass 0.98 kg at rest on the floor which has u=.1 if the collision is completely inelastic find the distance in meter moved by the block

Answer 1

Answer:

Mass of bullet = m =

1000

20

kg

Mass of system after collision = m

′

=

1000

20+980

kg

final velocity after collision = v'

a)

2

1

m

′

v

′2

=

2

1

kx

2

v' =

m

′

k

x

20+980

100×1000

10cms

−1

=100cms

−1

=1ms

−1

b)Using conservation of momentum,

mv=m'v'

1000

20

v=

1000

980+20

×1

v=50ms

−1

c) Kinetic energy lost due to collision =

2

1

m

′

v

′2

−

2

1

mv

2

=-24.5 J

Explanation:

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Answer 2

The final answer would be -24.5 J kinetic energy lost.

Given,

Mas of the bullet= 20g

Speed of the bullet= 100m/s

Mass of the wooden block= 0.98kg

To Find,

The total kinetic energy lost due to the collision

Solution,

Mass of the bullet could be expressed as m

                                                                          $= \frac{20}{1000} kg$

Therefore, the entire mass after collision = m'

                                                                     $= \frac{20+980}{1000} kg$

Let us consider the final velocity to be v'

Therefore using the formula  $\frac{1}{2} m'v^{2} = \frac{1}{2}kx^{2}$, the value of v'

                        = $\sqrt{\frac{k}{m'} } x$ = 100$= 100cms^{-1} = 1ms^{-1}$

By using the formula of Conservation of Momentum, mv=m'v'

                                                     $\frac{20}{1000} v = \frac{980+20}{1000}$

therefore, $v= 50ms^{-1}$

Now, kinetic energy amount lost due to the collision $= \frac{1}{2} mv^{2}$

                                                                                        = -24.5 J

Hence, the total amount of kinetic energy lost due to collision will be - 24.5J.

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