Momentum: 1D collisions A car has a mass of 1850kg. A truck was travelling at 65.1km/hr just before impacting the stationary car from directly behind. After the car and truck lock together, they travelled at 26.2km/hr. The mass of the truck is?
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To find the mass of the truck, we can use the principle of conservation of momentum.
Let's denote:
- ( m_c ) as the mass of the car (1850 kg)
- ( m_t ) as the mass of the truck (unknown)
- ( v_{c,i} ) as the initial velocity of the car (0 m/s, since it's stationary)
- ( v_{t,i} ) as the initial velocity of the truck (65.1 km/hr converted to m/s)
- ( v_{f} ) as the final velocity of the combined system (26.2 km/hr converted to m/s)
By conservation of momentum:
[ m_c \cdot v_{c,i} + m_t \cdot v_{t,i} = (m_c + m_t) \cdot v_f ]
Solving for ( m_t ):
[ m_t = \frac{(m_c \cdot v_{c,i} + m_t \cdot v_{t,i})}{v_f} - m_c ]
Substitute the given values and solve for ( m_t ).
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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