Question

A bullet of mass 20 g strikes a block of mass 980 g
with a velocity v and is embedded in it. The block
is in contact of a spring whose force-constant is
100 N/m. After the collision the spring is compressed
up to 10 cm. Find (a) magnitude of the velocity v of the bullet​

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

Answer 2

This will help you.

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Thank You❤