A projectile of mass `50 kg` is shot vertically upwards with n initial velocity of `100m//s`. After `5s`, it explodes into two fragments, one of (1st fragment) which having a mass of `20 kg` travels vertically up with a velovity of `150m//s (g = 10 m//s^(2))`. Bases on the above paragraph answer the following questions. What is the magnitude and direction of velocity of the 2nd fragments just after explosion isA. `(50)/(3)m//s` (up)B. `50m//s` (down)C. `50m//s`D. `(50)/(3)m//s` (down)

A projectile of mass `50 kg` is shot vertically upwards with n initial

1.	A projectile of mass `50 kg` is shot vertically upwards with n initial velocity of `100m//s`. After `5s`, it explodes into two fragments, one of (1st fragment) which having a mass of `20 kg` travels vertically up with a velovity of `150m//s (g = 10 m//s^(2))`. Bases on the above paragraph answer the following questions. What is the magnitude and direction of velocity of the 2nd fragments just after explosion isA. `(50)/(3)m//s` (up)B. `50m//s` (down)C. `50m//s`D. `(50)/(3)m//s` (down)
Answer» Correct Answer - D Velocity of projectile after `5s` is `= 50 m//s` Total momentum before explosion `=` Total momentum after explosion. `m xx 50 = m_(1)v_(1)+m_(2)v_(2)` `50 xx 50 = 20 xx 150 - 30 xx v_(2)` `v_(2) = (50)/(3)m//s` (velocity of 2nd fragment just after explosion) velocity of `2^(nd)` fragment `3s` after the explosion is `= (50)/(3) + 10 xx 3 = (140)/(3)`, momentum of `2^(nd)` fragement `3s` after the explosion is `(140)/(3) xx 30` Velocity of `1^(st)` one `3s` after the explosion is `= 150-10 xx 3 = 120 m//s` Sum of momentum of two fragment `3s` after the explosion `= 20 xx 120 - (30 xx 140)/(3)=2400-1400 = 1000 kg-m//s`

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