Systems of Particles & Linear Momentum

Only one object's momentum changes because forces cannot be shared.

Momenta changes are independent for both objects based on their respective masses only.

Both objects' momenta increase regardless of direction due to conservation of energy.

Their momenta change by equal amounts but in opposite directions due to Newton's third law.

There's no effect on impulse as only force magnitude affects change in momentum irrespective of action duration.

Time duration inversely affects impulse such that doubling application time halves imparted impulsive action for same force amount.

It decreases overall impulse as longer application leads to diminishing returns on imparted changes in momentum.

Increasing time duration increases impulse since it's directly proportional to time interval for constant force magnitude.

The momentum will only change if mass is added or removed from the cart.

It will cause a proportional change in acceleration instead of momentum.

The momentum remains unchanged because there's no friction.

The momentum will change by an amount equal to the impulse.

The system would register a higher speed for the projectile due to increased transfer of momentum between the block and projectile.

The system would register a slower speed for the projectile due to increased energy loss in the strings and erect suspension point.

The system would register a lesser speed for the projectile due to a conserved impulse during a shorter collision time.

There is no change in the registered speed for the projectile since the collision is inelastic regardless of string rigidity.

The moving cart reverses direction post-collision due to an elastic collision causing inversion of velocities.

Both carts move forward at half of the original cart's initial velocity due to conservation of momentum.

The moving cart comes to rest while transferring all its velocity to the other cart.

The moving cart continues forward at its initial velocity because kinetic energy is conserved in elastic collisions.

It equals the ball’s initial momentum horizontally.

It cannot be determined without knowing the mass of the wall.

It is twice the ball’s initial momentum horizontally.

It equals zero because speed does not change.

$\sqrt{\frac{g \cdot h}{M+m}}$

$\frac{g \cdot h \cdot (M-m)}{M}$

$\sqrt{\frac{g \cdot h \cdot (M+m)}{M}}$

$\sqrt{g \cdot h}$

The change in momentum remains constant at $J$ since impulse is defined as the product of force and time, which are independent choices here.

Change in momentum reduces to $\frac{J}{4}$ because longer duration leads to spreading out force application, reducing efficiency.

The change in momentum doubles to $2J$ because longer interaction time means larger impact on particle.

The change in momentum is four times greater or $4J$ as extending time multiplies the overall effect by same factor.

$v$ m/s

$0.5v$ m/s

$0$ m/s

$-v$ m/s

No detectable motion occurs since impulses do not necessarily result in movement.

The ball spins about its axis without gaining any linear momentum.

The ball gains linear momentum equal to the magnitude of that impulse.

The ball gains vertical motion due solely to gravitational pull.