6. Name: __ Date: __ __ Calculating Impulse and Change in Momentum p=m(v) Delta meanschangein p=momentum Delta p=m(v_(f)-v_(1)) m=mass F(t)=m(v_(f)-v_(i)) v_(f)=velocity final v_(i)=velocity initial or beginning F=forcet=time Ap 1. m What is the change in momentum of a 1000 kg car moving from 1m/s to 8m/s v_(f) Delta p=m(v_(f)-v_(i)) Delta p=1000(8-1) Delta p=1000(7) Delta p=underline ( ) 2. A tennis ball may leave a top player's racket on the serve with a speed of 65.0m/s If m Ap the ball's mass is 0.0600 kg, what is the change of momentum of the ball if it started from rest? How much force is required to stop a 60 kg person traveling at 30m/s during a time of: a. 5.0 seconds __ b. 0.50 seconds __ c. 0.05 seconds __ 4. A 1200-kg car is brought from 25m/s to 10m/s over a ti time period with a force c of -600N Determine the time it took for this change in momentum. A 0.10-kg hockey puck decreases its speed from 40m/s to 0m/s in 0.025 seconds. Determine the force that it experiences. A force of 6.0 N acts on a 4.0 kg object for 10.0s. What is the object's change in momentum?

1. The change in momentum of a 1000 kg car moving from 1 m/s to 8 m/s is calculated as follows:<br />$\Delta p = m(v_f - v_i) = 1000(8 - 1) = 1000(7) = 7000 kg \cdot m/s$<br /><br />2. The change in momentum of the tennis ball is calculated as follows:<br />$\Delta p = m(v_f - v_i) = 0.0600(65.0 - 0) = 0.0600(65.0) = 3.9 kg \cdot m/s$<br /><br />3. The force required to stop a 60 kg person traveling at 30 m/s during a time of:<br />a. 5.0 seconds: $F = \frac{\Delta p}{t} = \frac{60(30)}{5.0} = 360 N$<br />b. 0.50 seconds: $F = \frac{\Delta p}{t} = \frac{60(30)}{0.50} = 3600 N$<br />c. 0.05 seconds: $F = \frac{\Delta p}{t} = \frac{60(30)}{0.05} = 36000 N$<br /><br />4. The time it took for the 1200-kg car to change its momentum from 25 m/s to 10 m/s with a force of -600 N is calculated as follows:<br />$\Delta p = m(v_f - v_i) = 1200(10 - 25) = 1200(-15) = -18000 kg \cdot m/s$<br />$F = \frac{\Delta p}{t} \Rightarrow t = \frac{\Delta p}{F} = \frac{-18000}{-600} = 30 s$<br /><br />5. The force experienced by the 0.10-kg hockey puck as it decreases its speed from 40 m/s to 0 m/s in 0.025 seconds is calculated as follows:<br />$\Delta p = m(v_f - v_i) = 0.10(0 - 40) = 0.10(-40) = -4 kg \cdot m/s$<br />$F = \frac{\Delta p}{t} = \frac{-4}{0.025} = -160 N$<br /><br />6. The object's change in momentum when a force of 6.0 N acts on a 4.0 kg object for 10.0 s is calculated as follows:<br />$\Delta p = F \cdot t = 6.0 \cdot 10.0 = 60 kg \cdot m/s$