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Multiple choice physics work, energy and power collision of two rigid bodies energy and collisions understanding collisions

A sphere $A$ moving with speed $u$ and rotating with an angular velocity $\omega$ makes a head-on elastic collision with an identical stationary sphere $B$. There is no friction between the surfaces of $A$ and $B$. Choose the correct alternative(s). Discard gravity.

  1. $A$ will stop moving but continue to rotate with an angular velocity $\omega$
  2. $A$ will come to rest and stop rotating
  3. $B$ will move with speed $u$ without rotating
  4. $B$ will move with speed $u$ and rotate with an angular velocity $\omega$.
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation
Let $m$ be the mass of sphere and $v _1$ and $v _2$ be the velocities of the spheres A and B respectively after the collision.
Applying conservation of linear momentum:        $P _i = P _f$
$m  u + 0 = mv _1  + mv _2           \implies v _1 + v _2 = u$      ........(1)
Also $\dfrac{v _2- v _1}{ u-0} = -1         \implies v _1 - v _2 = -u$  
On solving we get,       $v _1 = 0 $   and $v _2 = u$
Thus A will stop and B will move with $u$
Also as there is no friction between the A and B, thus there will be no torque. Hence angular velocities of the respective spheres must remains the same as it was initially.
So, A will continue to rotate with $w$ whereas B will not rotate.
Multiple choice physics work, energy and power collision of two rigid bodies energy and collisions understanding collisions

A 20 g ball is fired horizontally toward a 100 g ball that is hanging motionless from a 1.0-m-long string. The balls undergo a head-on, elastic collision, after which the 100 g ball swings out to a maximum angle of 50 degrees. Determine the initial speed of the 20 g ball.

  1. 8.4m/s

  2. 4.2m/s

  3. 2.1m/s

  4. 16.8m/s
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

At first we can find the final velocity of the $100\ g$ ball.

$ \dfrac { { m } _{ 2 }{ { v } _{ 2 } }^{ 2 } }{ 2 } =mgh={ m } _{ 2 }gl(1-cos\alpha )$

$ \implies { v } _{ 2 }=\sqrt { 2gl\left( 1-cos\alpha  \right)  } =\sqrt { 2*9.81*1.1(1-cos{ 50 }^{ o }) } =2.78m/s $

Now applying conservation of the momentum principle, we get:

$ { m } _{ 1 }{ V } _{ 1 }+{ m } _{ 2 }{ V } _{ 2 }={ m } _{ 1 }{ v } _{ 1 }+{ m } _{ 2 }{ v } _{ 2 }$

$ { v } _{ 1 }=\dfrac { { m } _{ 1 }{ V } _{ 1 }-{ m } _{ 2 }{ v } _{ 2 } }{ { m } _{ 1 } } ={ V } _{ 1 }-{ v } _{ 2 }\dfrac { { m } _{ 2 } }{ { m } _{ 1 } } $

Principle of kinetic energy conservation , we get 

$ \dfrac { 1 }{ 2 } { m } _{ 1 }{ { { V } _{ 1 } }^{ 2 } }+\dfrac { 1 }{ 2 } { m } _{ 2 }{ { V } _{ 2 } }^{ 2 }=\dfrac { 1 }{ 2 } { m } _{ 1 }{ { v } _{ 1 } }^{ 2 }+\dfrac { 1 }{ 2 } { m } _{ 2 }{ { v } _{ 2 } }^{ 2 }$


$ { v } _{ 1 }={ V } _{ 1 }-{ v } _{ 2 }\dfrac { { m } _{ 2 } }{ { m } _{ 1 } } \ { V } _{ 2 }=0$

$\implies { V } _{ 1 }=\dfrac { { { { { m } _{ 1 } }^{ 2 }v } _{ 1 } }^{ 2 }+{ m } _{ 2 }{ m } _{ 1 }{ { v } _{ 2 } }^{ 2 } }{ 2{ m } _{ 1 }{ v } _{ 2 }{ m } _{ 2 } } =\dfrac { v _{ 2 }\left( { m } _{ 2 }+{ m } _{ 1 } \right)  }{ 2{ m } _{ 1 } } $

$ \implies  \dfrac { 2.78\left( 0.1+0.024 \right)  }{ 2*0.024 } =8.4m/s$

Multiple choice physics electronic devices boolean algebra digital electronics and logic gates logic gates

Which of the following is logically equivalent to $(p\wedge q)$ ?

  1. $p\rightarrow q$
  2. $\sim p \, \wedge \sim q$
  3. $p\, \wedge \sim q$
  4. $\sim (p\rightarrow \sim q)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

By De Morgan's laws and logical implication rules, p and q is equivalent to not (p implies not q). p -> not q is equivalent to not p or not q. The negation is p and q.

Multiple choice physics electronic devices boolean algebra digital electronics and logic gates logic gates

The Boolean expression $P+\overline { P } Q$, where $P$ and $Q$ are the inputs of the logic circuit, represents  

  1. AND gate

  2. NAND gate

  3. NOT gate

  4. OR gate
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using Boolean algebra, the expression P + P_bar Q can be simplified using the distributive law. Since P + P_bar = 1, the expression simplifies to P + Q, which represents the logical OR operation. Therefore, the circuit is an OR gate.

Multiple choice physics electronic devices boolean algebra digital electronics and logic gates logic gates

In the binary number system, the number $100$ represents

  1. one

  2. three

  3. four

  4. hundred
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In binary number system only $0$'s and $1$'s are used to built the whole number system.
Hence,
$0 = 0$
$1 = 1$
Start back at $0$ again, but add $1$ on the left,
$2 = 10$
$3 = 11$
Start back at $0$ again, and add one to the number on the left, but that number is already at $1$, so it also goes back to $0$ and $1$ is added to the next position on the left. Hence,
$4 = 100$ .... and so on.