The-system-is-initially-at-origin-and-is-moving-with-a-velocity-3-m-s-k-A-force-120t-newton-i-acts-on-mass-m-2-where-t-is-time-in-seconds-The-man-throws-a-light-ball-at-the-instant-when

Question Number 22574 by Tinkutara last updated on 20/Oct/17

The system is initially at origin and is

moving with a velocity (3 m / s) \overset{\land}{k} . A

force (120 t newton) \overset{\land}{i} acts on mass m_{2},

where t is time in seconds .

The man throws a light ball, at the

instant when m_{1} starts slipping on m_{2},

with a velocity 5 m / s vertically up w . r . t

himself . Taking the masses of blocks

and man as 60 kg each and assuming

that the man never slips on m_{2}, find

(a) The time at which man throws the

ball and

(b) The coordinates of the point where

ball lands . Neglect the dimensions of

the system . (g = 10 m / s^{2})

Commented by Tinkutara last updated on 20/Oct/17

Answered by Sahib singh last updated on 21/Oct/17

i f m_{1} = m_{2} = m_{m a n}

v_{i n i t i a l} = 3 \overset{}{} k \hat{} m / s

μ = 0.2

\Rightarrow a_{m a x i m u m o f m_{2}} = \frac{F_{f r i c t i o n}}{m_{2}}

= \frac{(60 \times 0.2 \times 10) i \hat{}}{60} m / s^{2}

= 2 i \hat{} m / s^{2}

\Rightarrow w h e n F = 360 i \hat{} N

a s t o t a l m a s s i s 180 k g

\Rightarrow a t t = 3 s

a l o n g x a x i s

F = 120 t

\Rightarrow \frac{d p}{d t} = 120 t

\Rightarrow d p = 120 t d t

\Rightarrow \int_{0}^{v} d p = \int_{0}^{3} 120 t d t

\Rightarrow m Δ v = 60 \times 9

\Rightarrow Δ v = \frac{60 \times 9}{180} = 3 i \hat{} m / s

\Rightarrow v_{m a n} = 3 i \hat{}

&

\frac{d x}{d t} = \frac{t^{2}}{3}

\Rightarrow x_{a t t = 3} = 3

&

v_{z} = 3

\Rightarrow z_{a t t = 3} = 9

\Rightarrow {\vec{r}}_{b a l l i n i t i a l l y} = 3 i \hat{} + 9 k \hat{}

\Rightarrow v_{h o r i z o n r a l o f b a l l} = (3 i \hat{} + 3 k \hat{}) m / s

\Rightarrow∣ v_{h o r i z o n t a l} ∣ = 3 \sqrt{2}

\Rightarrow R a n g e = \frac{2 \times ∣ v_{v e r t i c a l} ∣ \times ∣ v_{h o r i z o n t a l} ∣}{g}

\Rightarrow R = \frac{2 \times 5 \times 3 \sqrt{2}}{10}

\Rightarrow R (i n m e t e r s) = 3 \sqrt{2}

\Rightarrow a s t h e h o r i z o n t a l v e l o c i t y

h a d d i r e c t i o n (\frac{3 i \hat{} + 3 k \hat{}}{3 \sqrt{2}})

\Rightarrow \vec{R} w i l l a l s o h a v e s a m e

d i r e c t i o n

\Rightarrow c h a n g e i n p o s i t i o n v e c t o r o f t h e

b a l l

{\vec{r}}_{f i n a l} - {\vec{r}}_{i n i t i a l} = (3 i \hat{} + 3 k \hat{}) m

\Rightarrow {\vec{r}}_{f i n a l} = (6 i \hat{} + 9 k \hat{}) m

Commented by Sahib singh last updated on 21/Oct/17

t h i s i s w r o n g b e c a u s e t h a t

e x t e r n a l f o r c e i s n o t

a c c e l e r a t i n g b l o c k m_{2} .

S t a t i c f r i c t i o n a l f o r c e i s

a c c e l e r a t i n g t h e b l o c k .

t h e r e g o r e, I s o l v e d f o r

Commented by Tinkutara last updated on 21/Oct/17

(3, 0, 3) is wrong as you can check in the

Section - H .

Commented by Sahib singh last updated on 21/Oct/17

o k . i w i l l c h e c k

Commented by Sahib singh last updated on 21/Oct/17

o o p s! i m a d e t w o m i s t a k e s .

w i l l r e c t i f y n o w .

Commented by Sahib singh last updated on 21/Oct/17

m a x i m u m a c c e l e r a t i o n

o f t h e b l o c k d u e t o l i m i t i n g

f r i c t i o n . A n d s i n c e m_{2}

i s a b o u t t o s l i p \Rightarrow t h e w h o l e

s y s t e m i s h a v i n g s a m e

a c c e l e r a t i o n a t t h i s i n s t a n t .

N o w w e k n o w t h e a c c e l e r a t i o n

a n d m a s s w e a l r e a d y k n o w

\Rightarrow w e c a n c a l c u l a t e t h e f o r c e

o n s y s t e m .

Commented by Tinkutara last updated on 21/Oct/17

Thank you Sir!

Commented by Sahib singh last updated on 21/Oct/17

n o w i t s f i n e .

Commented by Tinkutara last updated on 21/Oct/17

B u t s i n c e w e n e e d a c c e l e r a t i o n o f m_{2},

s o f o r c e \div m a s s o f m_{2} = a c c e l e r a t i o n o f

m_{2} . S o \frac{120 t}{60} = 2 \Rightarrow t = 1 s . W h y t h i s i s

w r o n g ?

Commented by Sahib singh last updated on 21/Oct/17

Y o u a r e w e l c o m e :)