Question

║⊕QUESTION⊕║
I think nature’s imagination is so much greater than man’s, she’s never going to let us relax”
↓
CLASS 11
↑
A shell of mass 1 kg is fired vertically upwards with a speed of 100 m/s. At the highest point, it explodes into two equal fragments. One fragment is found to go down with a speed of 25 m/s, what is the speed and direction of the second fragment just after the collision.

Answer 1

Concept:-As there is no external force acting on it,,,so momentum will not change.

so,we have to simply conserve the momentum

mass of each fragment=0.5kg

initial momentum=1×100=100kgm/s

final momentum=25×0.5+0.5×v (assuming the velocity of other fragment in same direction as of first)

so,from above statement

12.5+0.5v=100

0.5v=87.5

v=175m/s

hence the second fragment will also move in the same direction(as velocity comes positive) with magnitude 175m/s

Answer 2

$\huge{ \mathfrak{ \overline{ \underline{ \underline{ \red{ Answer☻}}}}}}$

ANSWER :

v=u+at

→v=uˆj−gtˆj=(100−10×5)ˆj=50ˆjm/s

m=1kg ,m1=0.4kg, m2=0.6kg, v1=25ms−1

m→v=m1¯v1=m2¯v2

1×50ˆj=−0.4×25ˆj+0.6→v2

⇒v2100ˆj = v2= 100ms−1

EXPLANATION :

A shell of mass

1 kg

is thrown vertically upwards with a speed of

100 m / s

. After 5 seconds, it explodes into two fragments. One fragment of mass

400 gm

is found to go down with a speed of

25 m / s

. What will happen to the second fragment just after the explosion ?

( g = 100 ms -1 )