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4-1.Newton's Laws of Motion
normal
In the system shown in the figure there is no friction anywhere. The block $C$ goes down by $a$ distance $x_0 = 10\, cm$ with respect to wedge $D$ when system is released from rest. The velocity of $A$ with respect to $B$ will be $(g = 10\, m/s^2)$ ......... $m/s$
A$0$
B$1$
C$2$
DNone of these
Solution
for Block C:
$20-2 T=m a_c$
$20-2 T=2 a_c \text { (1) }$
for Block $A:$
$T=1 \times a_a \text {-(2) }$
Now string same so,
$a_a=a_c$
Solving $(1)$ and $(2)$:
$20=4 a_c \Rightarrow a_c=5 m / s ^2=a_a$
Now for $A$ :
$v^2=u^2+2 a s$
$v^2=0+2 \times 5 \times 101100$
$V_A=1 m / s ^2 \text { duso } v_B=1 m / s ^2$
Now sinu they are approaching to each Other hence $v_A-v_B=1+1=2\,m / s$
$20-2 T=m a_c$
$20-2 T=2 a_c \text { (1) }$
for Block $A:$
$T=1 \times a_a \text {-(2) }$
Now string same so,
$a_a=a_c$
Solving $(1)$ and $(2)$:
$20=4 a_c \Rightarrow a_c=5 m / s ^2=a_a$
Now for $A$ :
$v^2=u^2+2 a s$
$v^2=0+2 \times 5 \times 101100$
$V_A=1 m / s ^2 \text { duso } v_B=1 m / s ^2$
Now sinu they are approaching to each Other hence $v_A-v_B=1+1=2\,m / s$
Standard 11
Physics
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