Conservation of Linear Momentum, Collisions

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1 Conseration of Linear Momentum, Collisions 1. 3 kg mass is moing with an initial elocity i. The mass collides with a 5 kg mass m, which is initially at rest. Find the final elocity of the masses after the collision if it is perfectly inelastic. ccording to the Law of Conseration of Momentum, Δ P = 0 m = ( m + m ) 1 1i 1 f m1 = m + m 1 1i 3kg = (10 m/s) 3kg + 5kg = 3.75 m/s f. hefty man of mass 150 kg jumps from a bridge onto a boat of mass 400 kg that is moing up rier at a speed of 5 m/s. How fast is the boat moing afterwards? This problem inoles a perfectly inelastic collision. long the rier direction (x axis), from the Law of Conseration of Momentum, we get m = ( m + m ) boat boat ( before) boat man boat ( after ) The elocity of the boat afterwards is then m 400kg = = (5m/s) = 3.64m/s kg boat boat ( after ) boat ( before) mboat mman 3. Two blocks are traeling toward each other. The first has a speed of 10 cm/sec and the second a speed of 60 cm/sec. fter the collision, the second block is moing with a speed of 0 cm/sec in a direction opposite to its initial elocity. If the mass of the first block is twice that of the second, determine the following: a. the elocity of the first block after collision; b. whether the collision was elastic or inelastic. Draw both 'before' and 'after' pictures and select a coordinate system as shown. efore V i V i

2 fter V f V f a. Since the surface is frictionless, the net force acting on the system is 0 N, and from Law of Conseration of Momentum P 1 i + P i = P + P f dding the x-components we hae m 1 1i + m i = m 1 1f + m f. Since m 1 = m we find that m + 1m = m + 1m 1i i 1f f + 1 = + 1 1i i 1f f (10 cm/s) + 1( 60 cm/s) = + 1(0 cm/s) = 60 cm/s = 30 cm/s The negatie sign ahead of the 30 means that after the collision 1 is moing to the left. b. The initial KE is gien by KE = (1/ ) ( m )( ) + (1/ ) ( m )( ) i 1 1i i = (1/ )( m )(10cm/s) + (1/ )( m )( 60cm/s) = (1/ )( ) m (cm/s) = 1900 m (cm/s) The final KE is gien by KE = (1/ )( m )( ) + (1/ )( m )( ) f 1 1f f = (1/ )( m )( 30cm/s) + (1/ )( m )(0cm/s) = (1/ )( ) m (cm/s) = 1100 m (cm/s) Since KE f is not equal to KE i, the collision is inelastic.

3 4. What is the recoil elocity of a 4.0 kg rifle that shoots a kg bullet at a speed of 80 m/s? The rifle and the bullet interact with each other during the process of the shot and all other external forces can be ignored with respect to that process. Therefore, the total momentum is consered. The initial momentum of the rifle-bullet system is 0 because the rifle and the bullet were at rest in the beginning. y the Law of Conseration of Momentum, 0 = m r r f + m b b f mb 0.050kg Hence, rf = bf = (80m/s)= 3.5m/s mr 4.0kg The negatie indicates the rifle recoils in the opposite direction the bullet went. 5. block with a mass of 00 g is sliding with a speed of 1 cm/sec on a smooth table. The block is making a head-on, elastic collision with a block of unknown mass that is initially not moing. fter the collision, the elocity of the 00 g block is 4 cm/sec in the same direction as its initial elocity. Determine the mass of the second block and its speed after the collision. a. This is a collision problem in 1-dimension. Draw both 'before' and 'after' pictures and select a coordinate system. efore V i fter V f V f x Since the surface is frictionless, and since no work is performed by either, then the net force acting on the system is 0, and we hae conseration of linear momentum: p + p = p + p 1f f 1i i Thus, adding the x-components we hae m m = m This gies (00g)(4cm/s) + m = (00g)(1cm/s) m f f f f i = (00g)(8cm/s) (1)

4 Since we hae unknowns, we need another equation. The collision is elastic. Hence, we also hae conseration of KE and write m m m m + = fter multiplying both sides of the equation by, m + m = m + m 1 1f f 1 1i i Substituting the alues and using equation (1) to replace m f, we get (00 g)(4 cm/s) + m = (00 g)(1 cm/s) + 0 m f f f = (00 g)(1 cm/s) (00 g)(4 cm/s) ( m ) = (00g)(144)(cm/s) (00g)(16)(cm/s) f f ((00g)(8cm/s)) = 16 cm/s f = (00g)(18)(cm/s) Hence, f = 16 cm/sec, and from equation (1) m f = (00 g)(8cm/s) m (16 cm/s) = (00 g)(8cm/s) (8cm/s) m = (00g) (16cm/s) m = 100g 6. cannonball explodes into two pieces at a height of h = 100 m when it has a horizontal elocity x =4 m/s. (Use g = 10 m/s.) The masses of the pieces are kg and 3 kg. The 3 kg piece falls ertically to the ground 4 s after the explosion. What distance does the other piece fall? The momentum is consered during the explosion. We are using capital V to designate the elocity of the cannonball before the explosion, and small for the elocities after the explosion. ( m + m ) V = m + m 1 x 1 x ( m + m ) V = m + m 1 y 1 y m 1 = kg, m = 3 kg, V x = 4 m/s, V y = 0, x =0. The 3 kg piece falls from the height of 100 m in 4 s, so

5 g 0 = h+ yt t y y (10 m/s ) 0 = 100 m + y (4s) (4s) 0 = 100 m + (4s) 80 m (80 100) m = = 5m/s 4s Now we can calculate the elocity components for m 1, which is the kg body using ( m + m ) V = m + m 1 x 1 x (5 kg)(4 m/s) = ( kg) + (3kg)(0 m/s) (5 kg)(4 m/s) = ( kg) = (5kg) (4m/s) (kg) = 60m/s ( m + m ) V = m + m 1 y 1 y (5kg)(0m/s) = (kg) + (3kg)( 5m/s) 0 = (kg) (15kg m/s) = 7.5 m/s Its motion is a projectile motion. The time it takes for the piece to hit the ground is gien by d = d + t+ at = 100 m + (7.5m/s) t (10 m/s ) t 1 Using the quadratic formula to sole for t and taking the positie solution t 1 = 5.3 s. During this time interal, it traels Δ x= 1 t1 = (60m/s)(5.3s)=318m. Rounding to two digits of accuracy we hae Δ x = 30m. x

CJ57.P.003 REASONING AND SOLUTION According to the impulse-momentum theorem (see Equation 7.4), F t = mv

CJ57.P.003 REASONING AND SOLUTION According to the impulse-momentum theorem (see Equation 7.4), F t = mv Solution to HW#7 CJ57.CQ.003. RASONNG AND SOLUTON a. Yes. Momentum is a ector, and the two objects hae the same momentum. This means that the direction o each object s momentum is the same. Momentum is