3pt3pt 3pt3pt0pt 1.5pt3pt3pt Honor Phyic Impule-Momentum Theorem Spring, 2017 Intruction: Complete the following workheet. Show all of you work. Name: Anwer Key Mr. Leonard 1. A 0.500 kg ball i dropped from ret at a point 1.20 m above the floor. The ball rebound traight upward to a height of 0.700 m. a) What wa the magnitude and direction of the impule that the floor exert on the ball, when the ball hit the ground? Hint: Ue energy or kinematic to find the velocity of the ball before and after hitting the floor. Then ue the Impule-Momentum Theorem.) The known and unknown variable are: v 0 = 0 m m = 0.5 kg h h 0 = 1.2 m f = 0.7 m J =? Since there i no mention of in the problem, we need to ue the impule-momentum theorem to olve thi problem. However, in order to ue thi theorem, we will need to determine the velocity of the ball jut before and jut after it hit the ground. Since the ball doe not return to it initial height, we can ay that ome of the ball energy i lot when it hit the floor. However, energy i conerved before the ball hit the floor. Moreover, energy i alo conerved after the ball bounce off the floor. Uing conervation of energy, we can eaily calculate the velocity of the ball before, v 0, and after, v f, it hit the floor. E 0 = E f mgh 0 = 1 2 mv2 0 v0 2 = 2gh 0 v 0 = 2gh 0 E 0 = E f 1 2 mv2 f = mgh f vf 2 = 2gh f v f = 2gh f = 4.85 m = 3.7 m Note that v 0 i negative becaue the ball i initially moving downward. Plugging thee reult into the impule-momentum theorem, we get: J = p = p f p 0 = mv f v 0 ) = 0.5 kg) 3.7 m 4.85 m )) = 4.28 kg m b) If the ball i in contact with the floor for 5 10 3 ) econd, what average force did the floor exert on the ball? Recall that force i given by the equation: F = p/. Plugging our value into thi give: F = p = 4.28 kg m 5 10 3 ) = 855 N
Honor Phyic/Impule-Momentum Tet Review Page 2 of 7 Name: Anwer Key 2. A firehoe pray 160 kg of water ever econd, at a peed of 25 m. What force would you need to apply to the end of the firehoe to hold it in place? When a firehoe eject water, it generate a thrut force. Recall that we can derive the formula from F = p. Becaue the hoe change the momentum of the water it eject by p = m v, and the hoe m v doe thi for a ma m during a time, the momentum i changing at a rate given of: or ) m v, where m/ i the rate that water i ejected from the hoe in thi cae 160 kg/). So, ) m F = v. Plugging value into thi formula give: ) m F = v = 160 kg ) 25 m ) = 4000 N
Honor Phyic/Impule-Momentum Tet Review Page 3 of 7 Name: Anwer Key 3. A 0.50 kg ball i traveling to the right with a velocity of +10.0 m. The ball collide elatically with a econd ball, which ha a ma of 1.5 kg and i alo traveling to the right with a peed of 4.0 m. Calculate the final velocitie of both ball, auming a head-on colliion. The known and unknown variable are: m 1 = 0.50 kg m 2 = 1.5 kg v 1 = 10.0 m v 2 = 4.0 m v 1 =? v 2 =? During elatic colliion, both object move with different final velocitie. A a reult there are two unknown v 1 and v 2). We alo have two equation conervation of momentum and conervation of energy). Recall that we can rewrite conervation of energy and conervation of momentum a: v 1 v 2 = v 2 v 1 m 1 v 1 + m 2 v 2 = m 1 v 1 + m 2 v 2 Plugging our known value into thee two equation give: 6 = v 2 v 1 1) 11 = 0.5v 1 + 1.5v 2 2) Where the unit have been dropped for convenience and readability. When faced with two equation and two unknown, we firt pick one equation and olve for one of the unknown. In thi cae, I ll ue equation 1) to olve for v 1. Plugging thi into equation 2) allow u to olve for v 2. 6 = v 2 v 1 v 1 = v 2 6 3) 11 = 0.5v 2 6) + 1.5v 2 11 = 0.5v 2 3 + 1.5v 2 14 = 2v 2 v 2 = 7 m Finally, we can olve for v 1 by plugging thi reult into equation 3): v 1 = 7 6 = 1 m So, v 1 = 1 m and v 2 = 7 m.
Honor Phyic/Impule-Momentum Tet Review Page 4 of 7 Name: Anwer Key 4. A father ha a ma of 87 kg and i kating along the ice. He ee hi 22 kg on tanding on the ice. The father kate by hi on, and pick him up a he pae by. After grabbing hi on, the father kate away with a peed of 2.4 m. Ignoring friction, how fat wa the father moving before he grabbed hi on? The known and unknown variable are: m F = 87 kg v f = 2.4 m m = 22 kg v F,0 =? Thi problem involve a perfectly inelatic colliion. An inelatic colliion occur whenever the two object tick together after a colliion. In inelatic colliion, momentum i conerved but energy i not. Applying conervation of linear momentum, we get: m F v F,0 = m F + m b )v ) f mf + m b v F,0 = v f = 3 m m F 5. A 10 g bullet i fired horizontally at a 5 kg wooden target which it on top of a 1 m tall platform. The bullet travel at a peed of 200 m. The bullet hit the target and get embedded inide it. A a reult, the target flie off of the platform, and fall to the ground. How far, horizontally, doe the target travel before hitting the ground? m b = 0.01 kg y = 1 m m t = 5 kg x =? Firt, we need to calculate the peed that the target i moving immediately after the bullet get embedded in it. A alway, the momentum immediately before the colliion i equal to the momentum immediately after the colliion. Uing thi, we can calculate the velocity of the target after the bullet hit it: m b + m t ) v f = m b v 0 v f = mb = 0.4 m m b + m t After the bullet collide with the target, the target i in free-fall. To find the ditance the target travel before hitting the ground, we need to calculate the amount of time it will take the target to fall 1 m. Uing the kinematic equation y = 1 2 a y) 2 + v 0,y, we ee that: ) v 0
Honor Phyic/Impule-Momentum Tet Review Page 5 of 7 Name: Anwer Key y = 1 2 a y) 2 + 0 v 0,y ) 2 = 2 y = a y 2 y a y 2 1 m) = 9.8 m 2 = 0.452 Since there i no acceleration in the horizontal direction, the horizontal ditance the target travel i given by x = v x. Therefore, x = v x = 0.4 m ) 0.452 ) = 0.181 m 6. An automobile ha a ma of 2100 kg and velocity of +17 m. The car collide with a tationary car whoe ma i 1900 kg. The car lock bumper and kid off together with the wheel locked. a) What i the velocity of the two car immediately after the colliion? The known and unknown variable are: m 1 = 2100 kg v 1 = 17 m m 2 = 1900 kg v f =? J =? v 2 = 0 m x =? µ k = 0.68 Since the car tick together, the colliion i perfectly inelatic. Thi mean that momentum i conerved and energy i not. Applying conervation of linear momentum to olve for v f, give: m 1 + m 2 )v f = m 1 v 1 m1 v f = m 1 + m 2 = ) v 1 2100 kg 2100 + 1900 kg = 8.93 m ) 17 m b) What wa the impule acting on the kidding car from the moment after the colliion until the car come to ret?
Honor Phyic/Impule-Momentum Tet Review Page 6 of 7 Name: Anwer Key After the colliion, friction caue the car to come to ret. The impule caued by friction i eaily calculated uing the Impule-Momentum Theorem. J = p = p f p 0 = 0 m 1 + m 2 )v f = 3.5710 4 ) kg m 7. A 50.0 kg kater i traveling due Eat at a contant peed of 3.00 m. A 70.0 kg kater i moving due South at 7.00 m. The two kater collide, and hold on to each other after the colliion. a) What i the final peed of the two kater? The known and unknown variable are: m 1 = 50 kg v 1 = 3 m Eat m 2 = 70 kg v 2 = 7 m South θ =? v f =? Since the two kater cling together after colliding, the colliion i perfectly inelatic. Thi mean that momentum i conerved, and energy i not. Since the two kater are not moving along a onedimenional line, we mut apply conervation of momentum along the x and y-direction. Applying conervation of momentum along the x-direction Eat-Wet) give: m 1 + m 2 )v f,x = m 1 v 1,x + m 2 v 0 ) 2,x m1 v f,x = v 1 m 1 + m 2 = 1.25 m Similarly, applying conervation of linear momentum along the y-direction give: m 1 + m 2 )v f,y = m 1 v 0 1,y + m 2 v ) 2,x m2 v f,x = v 2 m 1 + m 2 = 4.08 m Combining thee reult, we ee that: v f,x = 1.25 m and v f,y = 4.08 m. Calculating the magnitude, we get:
Honor Phyic/Impule-Momentum Tet Review Page 7 of 7 Name: Anwer Key v f = vx 2 + vy 2 = 1.25 m )2 + 4.08 m )2 = 4.27 m b) In that direction are they moving after the colliion? Uing trigonometry, we can determine the direction of the kater motion after the colliion: tanθ) = v y v x ) θ = tan 1 vy v x = 73 South of Eat