1. a body of mass 1 kg makes an elastic collision with another body at rest and continues to move in the original direction after collision with a velocity equal to 1/5 of its original velocity. find the mass of the second body.
a) $\frac{2}{3}$kg b) $\frac{3}{2}$kg c) $\frac{5}{2}$kg d) $\frac{2}{5}$kg
2. a ball of mass 200g rests on a vertical post of height 20 m. a bullet of mass 10 g, travelling in horizontal direction, hits the centre of the ball. after collision both travels independently. the ball hits the ground at a distance 30 m and the bullet at a distance of 120 m from the foot of the post. the value of initial velocity of the bullet will be ($g = 10$ m/s²)
a) 120 m/s b) 60 m/s c) 400m/s d) 360 m/s
3. a ball b of mass m is lying at rest on the top surface of a smooth horizontal table 5m high. another moving ball a of same mass makes an elastic collision with b and b slides off the table and strikes the floor at a horizontal distance of 10m from the table. then select the correct alternative(s) ($g = 10$m/s²)
a) the velocity of the ball a before collision is 5 m/s
b) the kinetic energy of the ball b at the time of striking the ground is ($m imes100$)j
c) the velocity of the ball a before collision is 30 m/s
d) the kinetic energy of the ball b at the time when it strikes the ground is ($50 imes m$)j
4. a object of mass $m_1$ collides with another object of mass $m_2$, which is at rest. after the collision the objects move with equal speeds in opposite direction. the ratio of the mass $m_2:m_1$ is: 18 march, 2021 s - ii
a) 2:1 b) 3:1 c) 1:2 d) 1:1
5. a body of mass 2 kg makes an elastic collision with a second body at rest and continues to move in the original direction but with one fourth of its original speed. what is the mass of the second body? 9 april, 2019 s - i
a) 1.0 kg b) 1.5 kg c) 1.8 kg d) 1.2 kg
6. a body of mass 2 kg moving with 20m/s makes an elastic collision with a second body at rest and continues to move in the original direction but with one fourth of its original speed. find the speed of second body.
a) 10m/s b) 15m/s c) 20m/s d) 25m/s
7. given below are two statements : one is labelled as assertion (a) and the other is labelled as reason (r)
assertion (a) : body ‘p’ having mass m moving with speed ‘u’ has head - on collision elastically with another body ‘q’ having mass ‘m’ initially at rest. if $mll m$, body ‘q’ will have a maximum speed equal to ‘2u’ after collision.
reason (r) : during elastic collision, the momentum and kinetic energy are both conserved. in the light of the above statements, choose the most appropriate answer from the options given below
a) both a and r are correct and r is the correct explanation of a 18 march, 2021 s - i
b) a is not correct but r is correct
c) a is correct but r is not correct
d) both a and r are correct but r is not the correct explanation of a

Question

1. a body of mass 1 kg makes an elastic collision with another body at rest and continues to move in the original direction after collision with a velocity equal to 1/5 of its original velocity. find the mass of the second body.
a) $\frac{2}{3}$kg b) $\frac{3}{2}$kg c) $\frac{5}{2}$kg d) $\frac{2}{5}$kg
2. a ball of mass 200g rests on a vertical post of height 20 m. a bullet of mass 10 g, travelling in horizontal direction, hits the centre of the ball. after collision both travels independently. the ball hits the ground at a distance 30 m and the bullet at a distance of 120 m from the foot of the post. the value of initial velocity of the bullet will be ($g = 10$ m/s²)
a) 120 m/s b) 60 m/s c) 400m/s d) 360 m/s
3. a ball b of mass m is lying at rest on the top surface of a smooth horizontal table 5m high. another moving ball a of same mass makes an elastic collision with b and b slides off the table and strikes the floor at a horizontal distance of 10m from the table. then select the correct alternative(s) ($g = 10$m/s²)
a) the velocity of the ball a before collision is 5 m/s
b) the kinetic energy of the ball b at the time of striking the ground is ($m	imes100$)j
c) the velocity of the ball a before collision is 30 m/s
d) the kinetic energy of the ball b at the time when it strikes the ground is ($50	imes m$)j
4. a object of mass $m_1$ collides with another object of mass $m_2$, which is at rest. after the collision the objects move with equal speeds in opposite direction. the ratio of the mass $m_2:m_1$ is: 18 march, 2021 s - ii
a) 2:1 b) 3:1 c) 1:2 d) 1:1
5. a body of mass 2 kg makes an elastic collision with a second body at rest and continues to move in the original direction but with one fourth of its original speed. what is the mass of the second body? 9 april, 2019 s - i
a) 1.0 kg b) 1.5 kg c) 1.8 kg d) 1.2 kg
6. a body of mass 2 kg moving with 20m/s makes an elastic collision with a second body at rest and continues to move in the original direction but with one fourth of its original speed. find the speed of second body.
a) 10m/s b) 15m/s c) 20m/s d) 25m/s
7. given below are two statements : one is labelled as assertion (a) and the other is labelled as reason (r)
assertion (a) : body ‘p’ having mass m moving with speed ‘u’ has head - on collision elastically with another body ‘q’ having mass ‘m’ initially at rest. if $mll m$, body ‘q’ will have a maximum speed equal to ‘2u’ after collision.
reason (r) : during elastic collision, the momentum and kinetic energy are both conserved. in the light of the above statements, choose the most appropriate answer from the options given below
a) both a and r are correct and r is the correct explanation of a 18 march, 2021 s - i
b) a is not correct but r is correct
c) a is correct but r is not correct
d) both a and r are correct but r is not the correct explanation of a

Accepted Answer

# Explanation:
## Question 1:
### Step1: Use elastic - collision formulas
For an elastic collision, conservation of momentum $m_1u_1 + m_2u_2=m_1v_1 + m_2v_2$ and conservation of kinetic energy $\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$. Here $m_1 = 1\ kg$, $u_2 = 0$, $v_1=\frac{1}{5}u_1$.
From conservation of momentum: $m_1u_1=m_1v_1 + m_2v_2$, so $u_1=\frac{1}{5}u_1 + m_2v_2$, then $m_2v_2=\frac{4}{5}u_1$.
From conservation of kinetic energy: $\frac{1}{2}m_1u_1^2=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$. Substitute $v_1 = \frac{1}{5}u_1$ into it: $\frac{1}{2}m_1u_1^2=\frac{1}{2}m_1(\frac{1}{5}u_1)^2+\frac{1}{2}m_2v_2^2$.
$\frac{1}{2}m_1u_1^2-\frac{1}{50}m_1u_1^2=\frac{1}{2}m_2v_2^2$, $\frac{25 - 1}{50}m_1u_1^2=\frac{1}{2}m_2v_2^2$, $\frac{24}{50}m_1u_1^2=\frac{1}{2}m_2v_2^2$.
Since $m_2v_2=\frac{4}{5}u_1$, then $v_2=\frac{4}{5m_2}u_1$. Substitute into the kinetic - energy equation: $\frac{24}{50}m_1u_1^2=\frac{1}{2}m_2(\frac{4}{5m_2}u_1)^2$.
$\frac{24}{50}m_1=\frac{8}{25m_2}$, $m_2=\frac{2}{3}\ kg$.
# Answer:
a) $\frac{2}{3}\ kg$

## Question 2:
### Step1: Analyze the motion of the ball and bullet after collision
For the ball, it is a horizontal - projectile motion. $h = 20\ m$, $x_1 = 30\ m$, $h=\frac{1}{2}gt^2$, so $t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times20}{10}} = 2\ s$. The horizontal velocity of the ball after collision $v_{ball}=\frac{x_1}{t}=\frac{30}{2}=15\ m/s$.
For the bullet, $x_2 = 120\ m$, $t = 2\ s$, and from conservation of momentum $m_{bullet}u=m_{bullet}v_{bullet}+m_{ball}v_{ball}$, $m_{bullet}=10\ g = 0.01\ kg$, $m_{ball}=200\ g = 0.2\ kg$.
The horizontal velocity of the bullet after collision $v_{bullet}=\frac{x_2}{t}=\frac{120}{2}=60\ m/s$.
From conservation of momentum $0.01u=0.01\times60 + 0.2\times15$, $0.01u=0.6 + 3$, $0.01u = 3.6$, $u = 360\ m/s$.
# Answer:
d) $360\ m/s$

## Question 3:
### Step1: Analyze the horizontal - projectile motion of ball B after collision
The height of the table $h = 5\ m$, $h=\frac{1}{2}gt^2$, $t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times5}{10}} = 1\ s$. The horizontal distance of ball B $x = 10\ m$, so the horizontal velocity of ball B after collision $v=\frac{x}{t}=10\ m/s$.
In an elastic collision between two equal - mass objects ($m_A=m_B = m$) with ball B initially at rest, the velocity of ball A before collision is equal to the velocity of ball B after collision. So the velocity of ball A before collision is $10\ m/s$.
The kinetic energy of ball B at the time of striking the ground $K.E.=\frac{1}{2}mv^2+mgh$, $v = 10\ m/s$, $h = 5\ m$, $K.E.=\frac{1}{2}m\times10^2+m\times10\times5=50m + 50m=(m\times100)J$.
# Answer:
b) The kinetic energy of the ball B at the time of striking the ground is $(m\times100)J$

## Question 4:
### Step1: Use conservation of momentum and kinetic energy for elastic collision
Conservation of momentum: $m_1u_1=m_1v_1 + m_2v_2$, since $v_1=-v_2$ (equal speeds in opposite directions), $m_1u_1=m_1(-v_2)+m_2v_2=(m_2 - m_1)v_2$.
Conservation of kinetic energy: $\frac{1}{2}m_1u_1^2=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$, $\frac{1}{2}m_1u_1^2=\frac{1}{2}m_1v_2^2+\frac{1}{2}m_2v_2^2=\frac{1}{2}(m_1 + m_2)v_2^2$.
From $m_1u_1=(m_2 - m_1)v_2$, $u_1=\frac{(m_2 - m_1)}{m_1}v_2$. Substitute into the kinetic - energy equation: $\frac{1}{2}m_1(\frac{(m_2 - m_1)}{m_1}v_2)^2=\frac{1}{2}(m_1 + m_2)v_2^2$.
$\frac{(m_2 - m_1)^2}{m_1}=m_1 + m_2$, $m_2^2-2m_1m_2+m_1^2=m_1^2+m_1m_2$, $m_2^2 - 3m_1m_2=0$, $m_2(m_2 - 3m_1)=0$. Since $m_2
eq0$, $m_2:m_1 = 3:1$.
# Answer:
b) $3:1$

## Question 5:
### Step1: Apply conservation of momentum and kinetic energy for elastic collision
Let $m_1 = 2\ kg$, $u_2 = 0$, $v_1=\frac{1}{4}u_1$.
Conservation of momentum: $m_1u_1=m_1v_1 + m_2v_2$, so $m_1u_1=m_1\times\frac{1}{4}u_1+m_2v_2$, $m_2v_2=\frac{3}{4}m_1u_1$.
Conservation of kinetic energy: $\frac{1}{2}m_1u_1^2=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$. Substitute $v_1=\frac{1}{4}u_1$ into it: $\frac{1}{2}m_1u_1^2=\frac{1}{2}m_1(\frac{1}{4}u_1)^2+\frac{1}{2}m_2v_2^2$.
$\frac{1}{2}m_1u_1^2-\frac{1}{32}m_1u_1^2=\frac{1}{2}m_2v_2^2$, $\frac{16 - 1}{32}m_1u_1^2=\frac{1}{2}m_2v_2^2$, $\frac{15}{32}m_1u_1^2=\frac{1}{2}m_2v_2^2$.
Since $m_2v_2=\frac{3}{4}m_1u_1$, then $v_2=\frac{3}{4m_2}m_1u_1$. Substitute into the kinetic - energy equation: $\frac{15}{32}m_1u_1^2=\frac{1}{2}m_2(\frac{3}{4m_2}m_1u_1)^2$.
$\frac{15}{32}m_1=\frac{9m_1^2}{32m_2}$, $m_2 = 1.2\ kg$.
# Answer:
d) $1.2\ kg$

## Question 6:
### Step1: Use conservation of momentum and kinetic energy for elastic collision
$m_1 = 2\ kg$, $u_1 = 20\ m/s$, $u_2 = 0$, $v_1=\frac{1}{4}u_1 = 5\ m/s$.
Conservation of momentum: $m_1u_1=m_1v_1 + m_2v_2$. First, find $m_2$ as in the previous problem. $m_2 = 1.2\ kg$ (from the previous calculation).
$2\times20=2\times5+1.2v_2$, $40 = 10+1.2v_2$, $1.2v_2=30$, $v_2 = 25\ m/s$.
# Answer:
d) $25\ m/s$

## Question 7:
### Step1: Recall the concept of elastic head - on collision
In an elastic head - on collision between a heavy body of mass $M$ moving with speed $u$ and a light body of mass $m$ ($m\ll M$) initially at rest, from conservation of momentum $Mu=Mu_1+mu_2$ and conservation of kinetic energy $\frac{1}{2}Mu^2=\frac{1}{2}Mu_1^2+\frac{1}{2}mu_2^2$.
Solving these equations for $u_2$ when $m\ll M$ gives $u_2 = 2u$. And during elastic collision, both momentum and kinetic energy are conserved. So both assertion (A) and reason (R) are correct and (R) is the correct explanation of (A).
# Answer:
a) Both A and R are correct and R is the correct explanation of A

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QUESTION IMAGE

Question

Explanation:

Question 1:

Step1: Use elastic - collision formulas

Answer:

Question 2:

Step1: Analyze the motion of the ball and bullet after collision