Chapter 7

Impulse and Linear Momentum

Younes Sina

Linear MomentumLinear Momentum is defined as the product of mass and linear velocity.Pushing an object to the right, results in a reaction to the left.A rifle (attached to a cart) fired to the left makes the rifle move tothe right as shown:

As the bullet leaves the rifle at velocity Vb to the left, the rifle and the cart move to the right at velocity Vr .

Mb Vb = Mr Vr

Momentum to the left = Momentum to the right

Example :A solid ball of mass M1 = 0.15 kg is rolling to the right at speed V1 = 4.0 m/s andanother ball of mass M2=0.35 kg is rolling to the left at V2 = 6.0m/s.Find(a) the momentum of each ball (b) the net momentum.

Solution: (a) M1V1 = (0.15 kg)(+4.0 m/s) = + 0.60 kg m/sM2V2 = (0.35 kg)(-6.0 m/s) = - 2.1 kg m/s(b) ΣMV = M1V1 + M2V2 = -1.5 kg m/s

The net momentum is to the left

V

Example :A baseball of mass 0.120 kg is served by a pitcher horizontally to theleft at 17 m/s and it returns to the right at 63 m/s after getting struckby a bat. Calculate the change in its momentum.

Solution:Δ(MV) = MVf - MVi

Δ(MV) = M ( Vf - Vi )= 0.120kg [ ( + 63 m /s) - ( - 17 m /s) ] = + 9.6 kg m/s

Impulse:

Impulse ( I ) is the product of force and a time interval (FΔt).For example, if a grocery cart is pushed with a constant force of 44N to the left for 25 seconds, the impulse of the pusher on the cart isI = F ΔtI = (- 44N)(25s) = -1100 Ns

Note that Impulse and momentum have same units. The unit of momentum MV is (kg m/s).So is the unit of impulse: (Ns)=(kg m/s2)(s)=kg m/s

Equivalence of Impulse and Linear Momentum

FΔt = Δ(MV)F Δt = M Vf - M Vi

Impulse = Change in Momentum

Proof: Newton's 2nd law (for a single force):F = Ma

a =Δv/ΔtF = M Δv/Δt

Multiplying both side by Δt:FΔt = M Δv

FΔt = M (vf - vi)FΔt = Mvf - M vi

Example : A stationary train car of mass 12,000kg gets hit by another car moving to the right and is pushed with an average force of 4500Nfor a period of 4.2s. Find the final velocity of the stationary car.

Solution:Using the equivalence of Impulse and linear momentum:FΔt = M ( vf - vi )(+4500N)(4.2s) = (12,000kg)( vf - 0 )vf = +1.6 m/s ( to the right )

Momentum Change

Momentum Change = Impulse

Example: Wall exerts a force of 10,000 N.

The contact time is 0.01 s------------------------------Impulse = F t = 100 Ns

Δ (mv) = Change in momentum = final - initial

= 0 - mv

= - mvIgnoring negative sign: Δ(mv) = mv

Δ(mv) = Ftmv = Ft

Example :A 0.150-kg base ball is thrown horizontally to the left by a pitcher.Its velocity just before getting hit by the bat is 15 m/s to the left and after the strike becomes 45 m/s to the right. Find(a) the change in velocity Δv(b) the change in momentum MΔv(c) the impulse of the bat on the ball(d) the average force of the bat on the ball if the contact time is 0.020s.

Solution:(a) Δv = vf - vi = ( + 45 m/s ) - ( -15 m/s ) = + 60 m/s(The change in velocity, not the change in speed)(b) MΔv = (0.150 kg)( + 60. m/s) = + 9.0 kg m/s(c) According to the equivalence of impulse and linear momentum: FΔt = 9.0 kg m/s(d) FΔt = + 9.0 kg m/sF (0.020s) = + 9.0 kg m/sF = + 450N

Conservation of Linear Momentum

The total momentum before collision is : M1V1 + M2V2 (V1 and V2 are velocities before collision).

The total momentum after collision is : M1 u1 + M2 u2 (u1 and u2 are velocities after collision).

M1u1 + M2u2 = M1V1 + M2V2

Total momentum after collision = Total momentum before collision

M1u1 + M2u2 = M1V1 + M2V2

Example : A 1.00-kg toy car moving to the right at 1.40 m/s is hit from behind with a 0.500-kg piece of dough thrown horizontally also to the right at 3.60 m/s that causes the car and dough combination move faster. Calculate the speed of thecar-dough combo, knowing that the dough sticks to car.

Solution:Total momentum after collision =Total momentum before collision

McVc + MdVd = ( Mc + Md ) Vcd

(1.00 kg)(1.40 m/s) + (0.500 kg)(3.6 m/s) = (1.00 + 0.500) kg (Vcd)Vcd = 2.13 m/s

1 kg0.5 kg

Example :A 4.50-kg rifle is fixed on a 1.50-kg cart so that its barrel points horizontally to the right. The cart can roll with negligible friction and is initially at rest.The rifle is fired with a remote control device and shoots a 45.0 gram bullet to the right. As a result the rifle itself moves to the left at 2.50 m/s.Calculate the bullet exit speed.

Solution: Total momentum after collision must be equal to the total momentum before collision.Since before firing (or collision), both the bullet and rifle are at rest, total momentum before firing is zero. According to the law of conservation of linear momentum, the total momentum after firing must also be equal to zero as well:

(MbVb + MrVr)before firing = (MbVb + MrVr)after firing

(Note that Mr is not just the mass of rifle, it is the mass of rifle and cart).(0.045kg)( 0 )+ (4.50kg + 1.50kg)(0) = (0.045kg)(Vb)+ (4.50kg + 1.50kg)(- 2.50 m/s)0 + 0 = (0.045kg)(Vb) - 5.0 kg m/s15.0 kg m/s = 0.045 Vb

Vb = + 333 m/s [(+) means to the right]

Problems 1, 3, 5, and 7