The balls, having linear momenta $p_{1} = p \hat{i}$ and $p_{2} = - p \hat{i}$ , undergo a collision in free space. There is no extemal force acting on the balls. Let $p_{1}$ and $p_{2}$ be their final momenta. The following option(s) is (are) NOT ALLOWED for any non-zero value of $p, a_{1}, a_{2}, b_{1}, b_{2}, c_{1}$ and $c_{2}$ .

Question

Verified by Toppr

Answer 1

Since there is no net external force acting on the two body system, the net momentum of the system must remain conserved. The final momentum must not have any components in $j$ and $k$ directions as both momenta are in $i$ direction. Hence options $A$ and $D$ are not possible.

Answer 2

Since there is no net external force acting on the two body system, the net momentum of the system must remain conserved. The final momentum must not have any components in

j

and

k

directions as both momenta are in

i

direction. Hence options

A

and

D

are not possible.

$p_{1} = a_{1} \hat{i} + b_{1} \hat{j} + c_{1} \hat{k}$ ;

$p_{2} = a_{2} i + b_{2} \hat{j}$

$p_{1} = c_{1} \hat{k}$ ;

$p_{2} = c_{2} \hat{k}$

$p_{1} = a_{1} i + b_{1} \hat{j} + c_{1} \hat{k}$ ;

$p_{2} = a_{2} \hat{i} + b_{2} \hat{j} - c_{1} \hat{k}$

$p_{1} = a_{1} \hat{i} + b_{1} \hat{j}$ ;

$p_{2} = a_{2} i + b_{1} \hat{j}$

Since there is no net external force acting on the two body system, the net momentum of the system must remain conserved. The final momentum must not have any components in $j$ and $k$ directions as both momenta are in $i$ direction. Hence options $A$ and $D$ are not possible.

A ball is allowed to fall freely from a height of 3 metres on to a fixed plate. The successive rebound heights are $h_{1}, h_{2}, h_{3}, \dots$ lf the distance covered by the ball before coming to rest is y metres, find y. (Given that the coefficient of restitution is 0.5)

Two point masses $m_{1}$ and $m_{2}$ move with uniform velocity $ν_{1}$ and $ν_{2}$ respectively. Their initial positions are $r_{1}$ and $r_{2}$ respectively. Which is correct for the collision of two point masses ?

A proton of mass m collides elastically with a particle of mass M (which is originally at rest). The track of proton is deviated by $6 0^{o}$ . The struck particle (of mass M) follows a track at an angle of $3 0^{0}$ (with the incident proton direction). Find the value $(\frac{m}{M})$ .

lf the two bodies moving at right angles collide and their initial momenta are $P_{1}$ and $P_{2}$ , their resultant momentum after collison is :

Revise with Concepts

Collisions in Two Dimensions

Collisions - Problems L1

p​1​=a1​i^+b1​j^​+c1​k^; p​2​=a2​i+b2​j^​

p​1​=c1​k^; p​2​=c2​k^

p​1​=a1​i+b1​j^​+c1​k^; p​2​=a2​i^+b2​j^​−c1​k^

p​1​=a1​i^+b1​j^​; p​2​=a2​i+b1​j^​

Since there is no net external force acting on the two body system, the net momentum of the system must remain conserved. The final momentum must not have any components in j and k directions as both momenta are in i direction. Hence options A and D are not possible.

A ball is allowed to fall freely from a height of 3 metres on to a fixed plate. The successive rebound heights are h1​,h2​,h3​,… lf the distance covered by the ball before coming to rest is y metres, find y. (Given that the coefficient of restitution is 0.5)

Two point masses m1​ and m2​ move with uniform velocity ν1​ and ν2​ respectively. Their initial positions are r1​ and r2​ respectively. Which is correct for the collision of two point masses ?

A proton of mass m collides elastically with a particle of mass M (which is originally at rest). The track of proton is deviated by 60o. The struck particle (of mass M) follows a track at an angle of 300 (with the incident proton direction). Find the value (Mm​).

lf the two bodies moving at right angles collide and their initial momenta are P1​ and P2​, their resultant momentum after collison is :

Two particles having the position rˉ1​=(3i^+5j^​)m and rˉ2​=(−5i^−3j^​)m move with velocities Vˉ1​=(4i^+3j^​)m/s and Vˉ2​=(ai^+7j^​)m/s . If the particles collide, then value of a must be :

Revise with Concepts

Collisions in Two Dimensions

Collisions - Problems L1

$p_{1} = a_{1} \hat{i} + b_{1} \hat{j} + c_{1} \hat{k}$ ;

$p_{2} = a_{2} i + b_{2} \hat{j}$

$p_{1} = c_{1} \hat{k}$ ;

$p_{2} = c_{2} \hat{k}$

$p_{1} = a_{1} i + b_{1} \hat{j} + c_{1} \hat{k}$ ;

$p_{2} = a_{2} \hat{i} + b_{2} \hat{j} - c_{1} \hat{k}$

$p_{1} = a_{1} \hat{i} + b_{1} \hat{j}$ ;

$p_{2} = a_{2} i + b_{1} \hat{j}$

Since there is no net external force acting on the two body system, the net momentum of the system must remain conserved. The final momentum must not have any components in $j$ and $k$ directions as both momenta are in $i$ direction. Hence options $A$ and $D$ are not possible.

A ball is allowed to fall freely from a height of 3 metres on to a fixed plate. The successive rebound heights are $h_{1}, h_{2}, h_{3}, \dots$ lf the distance covered by the ball before coming to rest is y metres, find y. (Given that the coefficient of restitution is 0.5)

Two point masses $m_{1}$ and $m_{2}$ move with uniform velocity $ν_{1}$ and $ν_{2}$ respectively. Their initial positions are $r_{1}$ and $r_{2}$ respectively. Which is correct for the collision of two point masses ?

A proton of mass m collides elastically with a particle of mass M (which is originally at rest). The track of proton is deviated by $6 0^{o}$ . The struck particle (of mass M) follows a track at an angle of $3 0^{0}$ (with the incident proton direction). Find the value $(\frac{m}{M})$ .

lf the two bodies moving at right angles collide and their initial momenta are $P_{1}$ and $P_{2}$ , their resultant momentum after collison is :