A block c of mass is moving with velocity and collides elastica-Turito

Have a query? Ask a Tutor!!

Access to best educators and exclusive paper discussions

Can’t find what you’re looking for?

Our tutors are from top schools around the world, and they are waiting for you 24/7, anytime, anywhere.

Homework- One to One Solutions
Understand concepts to solve better
Get your doubts solved instantly

Physics-

General

Easy

Question

A block C of mass is moving with velocity and collides elastically with block of mass and connected to another block of mass through spring constant .What is if is compression of spring when velocity of is same ?

$\frac{{mv}_{0}^{2}}{X_{0}^{2}}$
$\frac{m v_{0}^{2}}{2 x_{0}^{2}}$
$\frac{3}{2} \frac{m v_{0}^{2}}{x_{0}^{2}}$
$\frac{2}{3} \frac{{mv}_{0}^{2}}{x_{0}^{2}}$

The correct answer is: $2 over 3 fraction numerator mv subscript 0 superscript 2 over denominator straight x subscript 0 superscript 2 end fraction$

Using conservation of linear momentum, we
have

Or
Using conservation of energy, we have

Where compression in the spring

Or

Book A Free Demo

Name*

Mobile*

+91

Email*

Grade*

I agree to get WhatsApp notifications & Marketing updates

Related Questions to study

If $fraction numerator 1 over denominator left parenthesis 1 plus 2 x right parenthesis open parentheses 1 minus x squared close parentheses end fraction equals fraction numerator A over denominator 1 plus 2 x end fraction plus fraction numerator B over denominator 1 plus x end fraction plus fraction numerator C over denominator 1 minus x end fraction$ then assending order of A,B,C

If $fraction numerator 1 over denominator left parenthesis 1 plus 2 x right parenthesis open parentheses 1 minus x squared close parentheses end fraction equals fraction numerator A over denominator 1 plus 2 x end fraction plus fraction numerator B over denominator 1 plus x end fraction plus fraction numerator C over denominator 1 minus x end fraction$ then assending order of A,B,C

The number of different seven digit numbers that can be written using only the three digits 1, 2 and 3 with the condition that the digit 2 occurs twice in each number is-

Complete step by step solution:
We are given the number of different seven-digit numbers that can be written using only three digits 1,2 and 3. Therefore,
Total number of Digits

= 7

We are given that the digit two occurs exactly twice in each number.
Thus, the digit two occurs twice in the seven digit number.
Now, we will find the number of ways of arrangement of the digit two in the seven digit number by using combination.
Total number of ways that the digit two occurs exactly twice in each number

=^{7} C_{2}

Now, the remaining five digits can be written using two digits 1 and 3 in

2 to the power of 5

ways.
We will now find the total number of seven digit number by multiplying the number of ways of arrangement in both the cases. Therefore
Total number of seven digit number =

C presuperscript 7 subscript 2 space cross times space 2 to the power of 5

Now by using the formula

C presuperscript n subscript r space equals space begin inline style fraction numerator n factorial over denominator r factorial space left parenthesis n minus r right parenthesis factorial end fraction end style end subscript

, we get
Total number of seven digit number =

fraction numerator 7 factorial over denominator left parenthesis 7 minus 2 right parenthesis factorial space 2 factorial end fraction cross times space 2 to the power of 5

We know that the factorial can be written by the formula n! = n

cross times

(n-1)! , so we get
Total number of seven digit number =

fraction numerator 7 space cross times 6 cross times 5 factorial over denominator 5 factorial space 2 factorial end fraction cross times 2 to the power of 5

Total number of seven digit number =

fraction numerator 7 space cross times 6 over denominator 2 factorial end fraction cross times 2 to the power of 5

Simplifying the expression, we get
Total number of seven digit number

= 7 \times 6 \times 2^{4}

= 7 \times 6 \times 2^{4}

Total number of seven digit number

= 672

Therefore, the number of different seven-digit numbers that can be written using only three digits 1,2 and 3 is 672.

The number of different seven digit numbers that can be written using only the three digits 1, 2 and 3 with the condition that the digit 2 occurs twice in each number is-

Complete step by step solution:
We are given the number of different seven-digit numbers that can be written using only three digits 1,2 and 3. Therefore,
Total number of Digits

= 7

We are given that the digit two occurs exactly twice in each number.
Thus, the digit two occurs twice in the seven digit number.
Now, we will find the number of ways of arrangement of the digit two in the seven digit number by using combination.
Total number of ways that the digit two occurs exactly twice in each number

=^{7} C_{2}

Now, the remaining five digits can be written using two digits 1 and 3 in

2 to the power of 5

ways.
We will now find the total number of seven digit number by multiplying the number of ways of arrangement in both the cases. Therefore
Total number of seven digit number =

C presuperscript 7 subscript 2 space cross times space 2 to the power of 5

Now by using the formula

C presuperscript n subscript r space equals space begin inline style fraction numerator n factorial over denominator r factorial space left parenthesis n minus r right parenthesis factorial end fraction end style end subscript

, we get
Total number of seven digit number =

fraction numerator 7 factorial over denominator left parenthesis 7 minus 2 right parenthesis factorial space 2 factorial end fraction cross times space 2 to the power of 5

We know that the factorial can be written by the formula n! = n

cross times

(n-1)! , so we get
Total number of seven digit number =

fraction numerator 7 space cross times 6 cross times 5 factorial over denominator 5 factorial space 2 factorial end fraction cross times 2 to the power of 5

Total number of seven digit number =

fraction numerator 7 space cross times 6 over denominator 2 factorial end fraction cross times 2 to the power of 5

Simplifying the expression, we get
Total number of seven digit number

= 7 \times 6 \times 2^{4}

= 7 \times 6 \times 2^{4}

Total number of seven digit number

= 672

Therefore, the number of different seven-digit numbers that can be written using only three digits 1,2 and 3 is 672.

The centre and radius of the circle $r equals c o s space theta minus s i n space theta$ are respectively

The centre and radius of the circle $r equals c o s space theta minus s i n space theta$ are respectively

The centre of the circle $r squared minus 2 r left parenthesis 3 c o s space theta plus 4 s i n space theta right parenthesis minus 24$ is

The centre of the circle $r squared minus 2 r left parenthesis 3 c o s space theta plus 4 s i n space theta right parenthesis minus 24$ is

The equation of the circle with centre at $open parentheses 1 comma 0 to the power of 0 end exponent close parentheses$ , which passes through the point $open parentheses 0 comma fraction numerator pi over denominator 2 end fraction close parentheses$ is

The equation of the circle with centre at $open parentheses 1 comma 0 to the power of 0 end exponent close parentheses$ , which passes through the point $open parentheses 0 comma fraction numerator pi over denominator 2 end fraction close parentheses$ is

The foot of the perpendicular from $left parenthesis negative 1 comma pi divided by 6 right parenthesis$ on the line $r left parenthesis 3 s i n space theta plus square root of 3 c o s space theta right parenthesis equals 3$ is

The foot of the perpendicular from $left parenthesis negative 1 comma pi divided by 6 right parenthesis$ on the line $r left parenthesis 3 s i n space theta plus square root of 3 c o s space theta right parenthesis equals 3$ is

The foot of the perpendicular from the pole on the line $r left parenthesis c o s space theta plus square root of 3 s i n space theta right parenthesis equals 2$ is

The foot of the perpendicular from the pole on the line $r left parenthesis c o s space theta plus square root of 3 s i n space theta right parenthesis equals 2$ is

The equation of the line parallel to $r left square bracket 3 C o s space theta plus 2 S i n space theta right square bracket equals 5$ and passing through $open parentheses 2 comma fraction numerator pi over denominator 2 end fraction close parentheses$ is

The equation of the line parallel to $r left square bracket 3 C o s space theta plus 2 S i n space theta right square bracket equals 5$ and passing through $open parentheses 2 comma fraction numerator pi over denominator 2 end fraction close parentheses$ is

The line passing through the points $open parentheses 2 comma fraction numerator pi over denominator 2 end fraction close parentheses$ , (3,0) is

The line passing through the points $open parentheses 2 comma fraction numerator pi over denominator 2 end fraction close parentheses$ , (3,0) is

Statement-I : If $fraction numerator 3 x plus 4 over denominator left parenthesis x plus 1 right parenthesis squared left parenthesis x minus 1 right parenthesis end fraction equals fraction numerator A over denominator x minus 1 end fraction plus fraction numerator B over denominator x plus 1 end fraction plus fraction numerator C over denominator left parenthesis x plus 1 right parenthesis squared end fraction$ then A= $7 over 4$
Statement-II : If $fraction numerator p x plus q over denominator left parenthesis 2 x minus 3 right parenthesis squared end fraction equals fraction numerator 1 over denominator 2 x minus 3 end fraction plus fraction numerator 3 over denominator left parenthesis 2 x minus 3 right parenthesis squared end fraction$ then $p equals 2 comma q equals 3$
Which of the above statements is true

Statement-I : If $fraction numerator 3 x plus 4 over denominator left parenthesis x plus 1 right parenthesis squared left parenthesis x minus 1 right parenthesis end fraction equals fraction numerator A over denominator x minus 1 end fraction plus fraction numerator B over denominator x plus 1 end fraction plus fraction numerator C over denominator left parenthesis x plus 1 right parenthesis squared end fraction$ then A= $7 over 4$
Statement-II : If $fraction numerator p x plus q over denominator left parenthesis 2 x minus 3 right parenthesis squared end fraction equals fraction numerator 1 over denominator 2 x minus 3 end fraction plus fraction numerator 3 over denominator left parenthesis 2 x minus 3 right parenthesis squared end fraction$ then $p equals 2 comma q equals 3$
Which of the above statements is true

If b > a , then the equation, (x - a) (x - b) - 1 = 0, has:

If b > a , then the equation, (x - a) (x - b) - 1 = 0, has:

If $alpha comma beta$ be that roots $4 x squared minus 16 x plus lambda equals 0$ where $lambda element of R$ , such that $1 less than alpha less than 2$ and $2 less than beta less than 3$ then the number of integral solutions of λ is

If $alpha comma beta$ be that roots $4 x squared minus 16 x plus lambda equals 0$ where $lambda element of R$ , such that $1 less than alpha less than 2$ and $2 less than beta less than 3$ then the number of integral solutions of λ is

If α,β then the equation $x squared minus 3 x plus 1 equals 0$ with roots $fraction numerator 1 over denominator alpha minus 2 end fraction comma fraction numerator 1 over denominator beta minus 2 end fraction$ will be

If α,β then the equation $x squared minus 3 x plus 1 equals 0$ with roots $fraction numerator 1 over denominator alpha minus 2 end fraction comma fraction numerator 1 over denominator beta minus 2 end fraction$ will be

The equation of the directrix of the conic $2 equals r left parenthesis 3 plus C o s invisible function application theta right parenthesis$ is

The equation of the directrix of the conic $2 equals r left parenthesis 3 plus C o s invisible function application theta right parenthesis$ is

The conic with length of latus rectum 6 and eccentricity is

The conic with length of latus rectum 6 and eccentricity is