if body is moving with the velocity of 100m/s and 10 balls per sec are thrown vith the velocity of 50 m/ s in the opposite direction in which the truck is moving . if each ball falls at a distance of 250 metres as the position of the truck was 10 seconds earlier . what is the position of 20 th ball thrown after the truck has gained the velocity of 100m/s and what is the it has taken to cover that distance at which 20 th ball falls
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Initially the body was at moving. The linear moments of the body is thus p=mu=0. The body direction is 250 metres due inertial force. As the external force acting on it zero.
p1=m1v1=(250g)×(20m/s) .
p2=m2v2=(250g )×v2
m1v1=m2v2
(250g) ×(20m/s) = (100g)×v2 or,v2 =540 m/so.
I HOPE THIS ANSWER TO YOUR HELP
p1=m1v1=(250g)×(20m/s) .
p2=m2v2=(250g )×v2
m1v1=m2v2
(250g) ×(20m/s) = (100g)×v2 or,v2 =540 m/so.
I HOPE THIS ANSWER TO YOUR HELP
buddy ita wrong
its**
according to me concept of conservation of momentum is not applicable here
it is write ya wrong
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