**Elastic Collisions, and Impulse**

Sports often contain many real-life
physics applications. This video

shows two basketball players in a collision.

Assuming that player 20’s mass was
90kg and player 13’s mass was 120kg.

Player 20 was traveling in the +x direction at 1.5m/s and
collided with player 13

traveling 2.0m/s in the –x direction (-2.0m/s). Assuming the
collision was perfectly

elastic, each player’s velocity (with direction) and impulse
(with direction after the

collision) can be calculated using the law of conservation
of momentum and the

impulse equation. (These calculations refer to the velocity
and impulse of each

player before player 20 fell.)The collision is assumed to be
elastic.

The law of conservation of momentum equation:

Since momentum is conserved in elastic collisions:

**[Equation 1]**

**M**

_{A}V_{A}+M_{B}V_{B}=M_{A}V^{’}_{A}+M_{B}V^{’}_{B}[Equation 2] V_{A}-V_{B}= -V^{’}_{A}+V^{’}_{B}
The axis of orientation is established and remains constant
throughout the problem:

positive velocities move in the (+x direction) to the right
and negative velocities

move in the (-x direction) to the left.

**M**

_{A}=90kg**M**

_{B}=120kg**V**

_{A}=1.5m/s**V**

_{B}= -2.0m/s
Unknowns:

**V**

^{’}_{A}=?**V**

^{’}_{B}=?
Plugging in the known initial velocities and masses to the
momentum equation:

**(90kg)(1.5m/s)+(120kg)(-2.0m/s)=(90kg)V**

^{’}_{A}+(120kg)V^{’}_{B}=-105m/s
using equation 2 to solve for V

^{’}_{B}in terms of V^{’}_{A}:**V**

^{’}_{B}=V^{’}_{A}+V_{A}-V_{B}**V**

^{’}_{B}=3.5m/s+V^{’}_{A}**-105m/s=(90kg)V**

^{’}_{A}+(120kg)(3.5m/s+V^{’}_{A})
solve for V

^{’}_{A}:**V**

^{’}_{A}= (-2.5m/s) (-x direction)
With 1 significant figure:

**V**

^{’}_{A}**≈ -3m/s (-x direction)**

Plugging -2.5m/s into equation 2 to solve for V

^{’}_{B}:**V**

^{’}_{B=}(3.5m/s)+(-2.5m/s)=1 m /s (+x direction)
With 1 significant figure:

**V**

^{’}_{B}**≈1m/s (+x direction)**

Now that the velocities are known, each player’s impulse can
be calculated:

Impulse on player #20:

**Impulse=M**

_{A}V^{’}_{A}-M_{A}V_{A}=the change in momentum for player 20
Plug in known values to solve for impulse:

**Impulse=(90kg)(-2.5m/s)-(90kg)(1.5m/s)= -360kg**

**Ÿm/s**

With one significant figure and direction, since impulse is
a vector:

**Impulse**

**≈ -400kg**

**Ÿm/s (-x direction)**

Impulse on player #13:

**Impulse=M**

_{B}V^{’}_{B}-M_{B}V_{B}=the change in momentum
Plug in known values to solve for impulse:

**Impulse=(120kg)(1m/s)-(120kg)(-2.0m/s)=360kg**

**Ÿm/s**

With one significant figure and direction, since impulse is
a vector:

**Impulse**

**≈ 400kg**

**Ÿm/s (+x direction)**

**Assuming the collision is perfectly elastic, momentum is conserved and the**

**players feel equal magnitudes of impulse in the opposite direction.**

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