Physics, asked by masoomabatool828, 1 month ago

A 10kg bat hits a 50g ball with 100N force in 1milisecond then what is impulse?​

Answers

Answered by bhakkiasree
1

Answer:

M=0.4 kg

h

1

=5m

h

2

=20m

g=10 m/s

2

velocity of ball after dropping 5m:u

velocity of ball after being hit by bat: v

u

2

=2gh

1

∴u=

2×10×5

u=10 m/s downwards

v

2

=2gh

2

∴v=

2×10×20

=20 m/s upwards

Change in momentum=impulse=m(v+u)

=0.4(10+20)

=0.4×30

=12 kg m/s

Force =100N=12/t

t=time of contact =0.12 sec.

Answered by vasundharakushwa33
0

Answer:

M=0.4 kg

h

1

=5m

h

2

=20m

g=10 m/s

2

velocity of ball after dropping 5m:u

velocity of ball after being hit by bat: v

u

2

=2gh

1

∴u=

2×10×5

u=10 m/s downwards

v

2

=2gh

2

∴v=

2×10×20

=20 m/s upwards

Change in momentum=impulse=m(v+u)

=0.4(10+20)

=0.4×30

=12 kg m/s

Force =100N=12/t

t=time of contact =0.12 sec.

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