Question

42% _ 1:07 PM D) If the collision is elastic, car 1 will never come to a stop after the collision E) If car 1 is much lighter than m2, and the collision is perfectly elastic, car 1 will continue heading to the right with nearly its original speed after the collision Answer: Subrmit All Answers 7.11pt A ball of mass 6.1 g with a speed of 21.3 m/s strikes a wall at an angle 11.0° and then rebounds with the same speed and angle. It is in contact with the wall for 40.0 ms. What is the magnitude of the impulse associated with the collision force? Answer Submit All Answers 8.[1pt] What is the average force exerted by the ball on the wall? Answer Submit Al Answers 9. 1pt] Suppose m is a certain mass and v is a certain speed v>0. A massm_1- 9m moving to the right with a velocity v 1+9v collides head-on with a second mass m-2-4m moving to the left with a velocity ?-2--4v. The two masses stick together to form a single mass moving with velocity v_f n V, where n is a number (no units). What is n?

Answer 1

Let the mass of the ball m = 6.1 g = 6.1 × 10^-3 kg

Let the velocity of the ball is V = 21.3 m/s

Angle with which ball strikes the wall is $\theta$ = 11°

Change in the momentum of the ball is equal to - 2 mvsin $\theta$ because the vertical components of momentum cancel each other and only horizontal components remain.

Impulse on the ball is equal to the change in its momentum

Impulse I = - 2 mV sin $\theta$

= - 2 × 6.1 × 10^-3 × 21.3 × sin11°

= - 12.2 × 21.3 × 10^-3 × 0.1908

= - 49.58 × 10^-3 kg m/s

B) Time for which ball is in contact with the wall is t = 40 ms = 40 × 10^-3 s

So average force = Impulse / time = I /t

= ( 49.58 × 10^-3 ) / ( 40 × 10^-3 )

= 1.23 N

C) Let the mass m1 = 9m and m2 = 4m

Velocity of m1 is v1 = 9v

Velocity of m2 is v2 = - 4v

Final velocity be v_f

As both the masses stick together they move with a common final velocity v_f and collision is inelastic.

Even though the collision is inelastic , we can apply law of conservation of momentum to find the final velocity.

m1 v1 + m2 v2 = ( m1 + m2 ) V

( 9 m × 9v ) + ( 4 m × - 4v ) = ( 9m + 4m ) v_f

81 mv - 16 mv = 13 m v_f

81 v - 16 v = 13 v_f

65 v = 13 v_f

v_f = ( 65/13 ) v = 5 v

We are given that v_f = nv . So the value of n = 5